# Thermodynamics Questions

### Questions

A $40\,{\rm g}$ block of ice is cooled to $-78{}^\circ {\rm C}$ and is then added to $560\, {\rm g}$ of water in an $80{\rm g}$ copper calorimeter at a temperature of $25{\rm {}^\circ C}$.  Determine the final temperature of the system consisting of the ice, water, and calorimeter.(${{\rm c}}_{ice}=0.5\,\frac{{\rm cal}}{{\rm g.}{\rm{}^\circ\!C}}{\rm \ ,\ }{{\rm c}}_{{\rm water}}=1\frac{{\rm cal}}{{\rm g.K}}\ ,\ c_{Cu}=0.0923\,\frac{{\rm cal}}{{\rm g.K}}$)

How much heat is needed to melt 20 ${\rm kg}$ of a solid material initially at $-10{}^\circ {\rm C}$? Given: $c_{material}=2500\ {\rm J/(kg\ C{}^\circ )}$ , $L_{material}=40\times {10}^4\ {\rm J/kg}$

The heat delivered or absorbed during a change in the temperature of an object is $Q=mc\Delta T$. If we have a phase change in the material, then the amount of heat absorbed or delivered is $Q=\pm mL$, where the plus sign is for melting process and the minus is for condensation. $L$ is called the latent heat and $c$ is the specific heat capacity of that substance.

The heat and temperature changes of this solid material is as follows
$solid\ -10{\rm{}^\circ\,\!C}\xrightarrow{Q_1}solid\ 0{\rm{}^\circ\,\!C}\xrightarrow{Q_2}fluid\ 0{\rm{}^\circ\,\!C}$
$Q_1=mc\Delta T=20\times 2500\times \left(0-\left(-10\right)\right)=50\times {10}^4\ {\rm J}$
$Q_2=+mL=20\times 40\times {10}^4=800\times {10}^4\ {\rm J}$
Thus, the total heat is the sum of them as
$Q_{total}=Q_1+Q_2=850\times {10}^4{\rm J}$

A person pours $330\, {\rm g}$ of water at $50{\rm{}^\circ\,\!C}$ into an $855\, {\rm g}$ aluminum container with an initial temperature of $10{\rm{}^\circ\,\!C}$. The specific heat of aluminum is $900\, {\rm J/(kg.K)}$ and that of the water is $4190\, {\rm J/(kg.K)}$. What is the final temperature of the system, assuming no heat is exchanged with the surroundings?

$2\, {\rm kg}$ of ice at $0{\rm{}^\circ\,\!C}$ is dropped into an insulated container with $5\, {\rm kg}$ of water at $80{\rm{}^\circ\,\!C}$.

(a) Find the final temperature.

(b) Find the entropy change of ice.

(c) Find the total change of entropy.

At $10{\rm{}^\circ\,\!C}$, $32\ {\rm g}$ of water is mixed with $430\ {\rm g}$ of ice at $-5\ {\rm{}^\circ\,\!C}$.
$(c_w=4190\ \left(\frac{{\rm J}}{{\rm kg.k}}\right)\ ,\ c_i=2090\ \left(\frac{{\rm J}}{{\rm kg.k}}\right)\ ,\ L_f=3.34\times {10}^5\frac{{\rm J}}{{\rm kg}})$

(a) What mass of water freezes?

(b) What is the change in entropy of the system?

A chunk of ice with a mass of $20\ {\rm kg}$ falls into the ocean and melts. Initially, and throughout the melting process, it is at temperature of $0{\rm{}^\circ\,\!C}$. The temperature of the very large body of ocean water around it is fixed at $5{\rm{}^\circ\,\!C}$.

(a) Calculate the change in the entropy of the ice, as it is converted to the liquid water, still at $0{\rm{}^\circ\,\!C}$.

(b) Calculate the change in the entropy of the ocean water as it supplies the heat to melt the ice.

(c) Calculate the total change in entropy, including its sign.

What is the entropy change of

(a) A $12\ {\rm g}$ ice cube that melts completely in a bucket of water whose temperature is just above the freezing point of water?

(b) A $5.0\ {\rm g}$ spoonful of water that evaporates completely on a hot plate whose temperature is slightly above the boiling point of water.

A student has just purchased a coffee. The cup has $0.3\ {\rm kg}$ of coffee at $70{\rm{}^\circ\,\!C}$. She adds $0.02\ {\rm kg}$ cream at $5{\rm{}^\circ\,\!C}$. In what follows assume coffee and cream have the same thermal properties as water.  ($c_w=4186\frac{{\rm J}}{{\rm kg.}{\rm{}^\circ\,\!C}}\ ,\ L_V=2.26\times {10}^6\frac{{\rm J}}{{\rm kg}}\ ,\ \ L_f=33.5\times {10}^4\frac{{\rm J}}{{\rm kg}}\ ,\ \rho=1\ {\rm g/}{{\rm cm}}^{{\rm 3}}$)

(a) What is the temperature of the coffee after adding cream? assume no heat is lost to the outside world at this point.

(b) For this process, is the entropy change of the universe less than, greater than, or equal to zero?

(c) In 1 minute, the coffee cools by $5{\rm{}^\circ\,\!C}$. In this part only, assume heat loss is entirely due to evaporation. What is the mass of the coffee lost in this 1 minute? You can also ignore the cream.

A $2\ {\rm kg}$ ice at $0\, {\rm ^\circ\, C}$ is dropped into an isolated container with $5\ {\rm kg}$ of water at $80{}^\circ\, {\rm C}$.

(a) Find the final temperature

(b) Find the entropy change of ice.

(c) Find the total change of entropy.
$(L_f=333\ {\rm kJ/K}, c_w=4186\ {\rm J/(kg.}{\rm{}^\circ\,\!C}{\rm )})$

A $100\ {\rm g}$ ice cube is mixed with a $200\ {{\rm cm}}^{{\rm 3}}$ of water at $70{\rm{}^\circ\,\!C}$ and allowed to reach thermal equilibrium in a thermally insulated container.

(a) What is the final temperature of the mixer?

(b) What is the change in entropy of the mixing?

(c) How many joules of energy must be added to the mixer to heat it to $100{\rm{}^\circ\,\!C}$ and vaporize it?

(d) Assuming the water vapor at $100{\rm{}^\circ\,\!C}$ is an ideal gas with density of $18\ {\rm g/mol}$, what volume will it occupy at a pressure of $1\times {10}^5{\rm Pa}$?

An ice-cube with mass $0.010\ {\rm kg}$ is removed from the freezer and placed into a glass of water. The temperature of the ice-cube before being put into the glass is $-8.5{\rm {}^\circ\, C}$, and the mass of the glass itself (with no water) is $0.100\ {\rm kg}$. The initial temperature of the water in the glass in $20.0{\rm {}^\circ\, C}$, and the final temperature of the water (after the ice cube has all melted and turned to water) is $16.9{\rm {}^\circ\, C}$.

(a) Calculate the entire heat gained by the ice-cube.

(b) By assuming that all of the heat gained by the ice-cube - what you have just calculated in part (a) - equals the heat lost by the glass and the water originally in the glass, calculate the mass of the water which was in the glass.
$(c_{ice}=2100\frac{{\rm J}}{{\rm kg.}{\rm{}^\circ\,\!C}}\ ,\ \ c_{glass}=840\ \frac{{\rm J}}{{\rm kg.}{\rm{}^\circ\,\!C}}\ ,\ \ L_f=3.33\times {10}^5\frac{{\rm J}}{{\rm kg}}\ \ ,\ \ c_w=4186\ \frac{{\rm J}}{{\rm kg.}{\rm{}^\circ\,\!C}})$

Molten lead of mass ${\rm 0.2\ kg}$ and temperature ${\rm 350{}^\circ\, C}$ is poured into a die cast made of aluminum. The mass of the cast is ${\rm 0.8\ kg}$ and the initial temperature is ${\rm 25{}^\circ\, C}$.

(a) How much heat is given off as the lead cools from liquid to solid at ${\rm 327{}^\circ\, C}$?~Hint: There are two parts to this calculation; ($i$) the heat given off in going from ${\rm 350{}^\circ\, C}$ to ${\rm 327{}^\circ\, C}$ in liquid state, and ($ii$) the change in phase at ${\rm 327{}^\circ\, C}$ going from liquid to solid

(b) What is the equilibrium temperature of the system assuming no heat loss?
$(L_{f-Pb}=2.5\times {10}^4\ {\rm J/kg} \quad, c_{Pb}=450\ {\rm J/(kg.}{\rm{}^\circ\,\!C}{\rm )}\quad, c_{Al}=900\ {\rm J/(kg.}{\rm{}^\circ\,\!C}{\rm )})$

A tank contains $600\ {{\rm m}}^{{\rm 3}}$ of neon at an absolute pressure of $1.01\times {10}^5{\rm \ Pa}$. The temperature of the gas is increased from $300\ {\rm K}$ to $302\ {\rm K}$. What is the increase in the internal energy of the neon?

The temperature of 5 moles of an ideal gas is 200 ${\rm K}$. how much work does the gas do in expanding isothermally to two times its initial volume? Given: ideal gas constant $R=8.31\ {\rm J/(mol\ K)\ }$,
${\ln 2\ }=0.7\ ,{\ln \frac{1}{2}\ }=-0.7\ ,{\ln 3\ }=1.1\ ,{\ln \frac{1}{3}\ }=-1.1\ ,{\ln 5\ }=1.6\ ,{\ln \frac{1}{5}\ }=-1.6$

By definition, the work done on the gas is $W_{on}=-P\Delta V$ and we know that $W_{on}=-W_{by}$ so the work done by the gas in an isothermal expanding is as follows
$W_{by}=\int{PdV}=\int^{V_f}_{V_i}{nRT\frac{dV}{V}}$
$W_{by}=nRT\,{\ln \frac{V_f}{V_i}\ }=5\left(8.31\right)\left(200\right){\ln \frac{2V_i}{V_i}\ }$
$W_{by}=8.31\times 1000\times 0.7=5817\ {\rm J\ \ }\Longrightarrow {\rm W\sim \ 6000\ J}$
In above we have used the ideal gas law $P=nRT/V$.

A gas is initially at (${\rm 20\ Pa,\ 8\ }{{\rm m}}^{{\rm 3}}$) and expands to (${\rm 27\ Pa,\ 12\ }{{\rm m}}^{{\rm 3}}$) .The minimum amount of pressure the gas can be under is ${\rm 9\ Pa}$, and the maximum pressure the gas can be under is ${\rm 40\ Pa}$. Find the minimum amount of work that can be done by the gas in going from its initial state to its final state.

One-mole of an ideal monoatomic gas has the $PV$ diagram as shown in the figure:

(a) Find the temperature at $A$ and $B$

(b) Find the works $W_{AB},W_{BC},W_{CD}$ done on the gas.

(c) Find the total heat exchanged from $A$ to $D$.

(a) Use the equation of state for an ideal gas, $PV=nRT$
$T_A=\frac{P_AV_A}{nR}=\frac{6000\times 1}{1\times 8.314}=721.6\ {\rm K}$
$T_D=\frac{P_DV_D}{nR}=\frac{6000\times 4}{1\times 8.314}=962.2\ {\rm K}$

(b) The work done on the system is given by $W_{on}=-P\Delta V$ or its magnitude by the area under the $P-V$ diagram. So
$W_{AB}=-P\left(V_B-V_A\right)=-6000\left(2-1\right)=-6000\ {\rm J}$
$\left|W_{BC}\right|=S_{trapezoid}=\frac{\left(2000+6000\right)\left(3-2\right)}{2}=4000\ {\rm J}$
Since $\Delta V_{BC}>0\$so due to $W=-P\Delta V$, the work done on the gas is negative i.e. $W_{BC}=-4000\ {\rm J}$
$W_{CD}=-P\Delta V_{CD}=-2000\left(4-3\right)=-2000\ {\rm J}$

(c) Use the first law of thermodynamics i.e. $\Delta U_{int}=W_{on}+Q_{in}$ and that the change in internal energy of an monoatomic ideal gas is ${\Delta U}_{int}=\frac{3}{2}nR\Delta T$, to find the desired heat
$\frac{3}{2}nR\left({{\rm T}}_{{\rm D}}{\rm -}{{\rm T}}_{{\rm A}}\right){\rm =}{\left({{\rm W}}_{{\rm AD}}\right)}_{on}+Q_{in}$
$W_{AD}=W_{AB}+W_{BC}+W_{CD}=-6000-4000-2000=-12000\ {\rm J}$
$\Rightarrow \ Q_{in}=\frac{3}{2}nR\left(T_D-T_A\right)-\left(-12000\right)=\frac{3}{2}\left(P_DV_D-P_AV_A\right)+12000$
$\Rightarrow Q_{in}=\frac{3}{2}\left(2000\times 4-6000\times 1\right)+12000=+15000\ {\rm J}$
Where in the third line we have replaced the $nRT$ by $PV$ to avoid uncertainty in the result. Since $Q_{in}>0$ thus the total heat absorbed by the gas.

A heat engine operates using the cycle shown in the figure. The working substance is $2\ {\rm mol}$ helium gas, which reaches a maximum temperature of $327{\rm{}^\circ\,\!C}$. Assume that the helium can be treated as an ideal gas. Process $bc$ is isothermal. The pressure in states $a$ and $c$ is ${10}^5{\rm Pa}$ and the pressure in state $b$ is $3\times {10}^5{\rm Pa}$.

(a) How much heat enters the gas and how much leaves the gas each cycle?

(b) How much work does the engine do each cycle, and what is its efficiency?

(c) Compare this engine efficiency with the maximum possible efficiency attainable with the hot and cold reservoirs used by this thermodynamic cycle.

A cylinder contains 7.1 moles of ideal gas, initially at a temperature of ${\rm 97{}^\circ\, C}$. The cylinder is provided with a frictionless piston, which maintains a constant pressure of $5.5\times {10}^5\ {\rm Pa}$ on the gas. The gas is cooled until its temperature has decreased to $27{\rm{}^\circ\,\!C}$. For the gas $C_V=12.24\ {\rm J/(mol.K)}$. The gas constant $R=8.314\ {\rm J/(mol.K)}$. Calculate:

(a) the work $W$ done by gas

(b) The net change in the internal energy, $\Delta U$, of the gas, and

(c) The heat transfer $Q$.

In a constant-volume process $1000\ {\rm J}$ of heat is transferred to $1.0\ {\rm mol}$ of a monoatomic ideal gas initially at $300\ {\rm K}$. Find: ($C_V=\frac{3}{2}R$)

(a) The work done on the gas

(b) The change in the internal energy of the gas

(c) The final temperature.

A $2\ {\rm mol}$ sample of an ideal diatomic gas is taken through the cycle shown in the figure. The process BC is an isothermal expansion. Calculate:

(a) The energy added to the gas by heat

(b) The energy exhausted from the gas by heat

(c) The net work done on the gas

(d) The efficiency of the cycle.

(a) One liter of a gas with $\gamma=1.3\$is at $273{\rm K}$ and $1.0\ {\rm atm}$ pressure. It is suddenly compressed adiabatically to half its original volume. Find its final pressure and temperature

(b) The gas is now cooled back to $273\ {\rm K}$ at constant pressure. What is its final volume?

One mole of an ideal diatomic gas goes from $a$ to $c$ along the diagonal path in the figure below. During the transition,

(a) What is the change in internal energy of the gas

(b) How much energy is added to the gas as heat?

(c) How much heat is required if the gas goes from $a$ to $c$ along the indirect path $abc$?

(d) What is total heat absorbed by the gas? And what is total work done on the gas?

(a) Use the definition of the internal energy as $U_{int}=nC_V\Delta T$.  Recall that for an ideal diatomic gas $C_V=\frac{5}{2}R$ so
$\Delta U_{int}=\frac{5}{2}nR\left(T_c-T_a\right)\xrightarrow{ideal\ gas\ law}\Delta U_{int}=\frac{5}{2}\left(P_cV_c-P_aV_a\right)$
$\Rightarrow \Delta U_{int}\left(a\to c\right)=\frac{5}{2}\left(2000\times 4-5000\times 2\right)=-5000\ {\rm J}$
Note: the change in internal energy of a system is independent of the path and only dependent on the initial and final temperature of the system, as the formulae shows this.
(b) See part (d), $Q\left(a\to c\right)=+2000\ {\rm J}$
(c) See part (d), $Q\left(a\to b\to c\right)=Q_1+Q_2=35000-30000=+5000\ {\rm J}$
(d) To solve this part, calculate work and heat of each process individually.
$a\to b\ :isobaric\ \left(\Delta P=0\right)$
$\ \ W_1=area\ under\ curve=\left(4-2\right)5000=10000\ {\rm J}$
\begin{align*}
Q_1 &=nC_P\Delta T_{ab}=n\ \left(C_V+R\right)\left(T_b-T_a\right)\\
&=\frac{7}{2}\left(P_bV_b-P_aV_a\right)=\frac{7}{2}\left(5000\right)\left(4-2\right)=35000\ {\rm J}
\end{align*}
$b\to c:isochoric\left(\Delta V=0\right),\ \ W_2=-P\Delta V=0$
\begin{align*}
Q_2 &=nC_V\Delta T_{bc}=\frac{3}{2}nR\left(T_c-T_b\right)\\
&=\frac{5}{2}\left(P_cV_c-P_bV_b\right)=\frac{5}{2}\left(2000-5000\right)\left(4\right)=-30000\ {\rm J}
\end{align*}
\begin{align*}
a\to c\ :W_3=area\ under\ curve(trapezoid)&=\frac{1}{2}(2000+5000)(4-2)\\
&=7000\ {\rm J}
\end{align*}
$Q_3=\Delta U_{int}\left(a\to c\right)-W_3=-5000-\left(-7000\right)=+2000\ {\rm J}$
Note: since in this case $\Delta V_{ac}>0$ and $\Delta V_{ab}>0\$so by definition of work done on the gas i.e. $W_{on}=-P\Delta V$, we have $W_3=-7000\ {\rm J}$ and $W_1=-10000\ {\rm J}$
Therefore the total heat added to the system, by convention (i.e. $Q_{in}>0$), is
$Q_{in}=Q_1=35000{\rm \ J}$
$W_{on}=-10000-7000=-17000\ {\rm J}$
Note: the area under $P-V$ diagram, gives the magnitude of the work done on the gas. To find the correct sign ($\pm$), we must use the formal definition of the work done on the gas that is $W_{on}=-P\Delta V$. If $\Delta V>0$, then $W_{on}<0$ and vice versa.
Note: the area of a trapezoid is as follows
$S=\frac{a+b}{2}h$

When $20.9\ {\rm J}$ was added as heat to a particular ideal gas, the volume of the gas changed from $50.0\ {{\rm cm}}^{{\rm 3}}$ to $100\ {{\rm cm}}^{{\rm 3}}$ while the pressure remained constant at $1.00\ {\rm atm}$.

(a) By how much did the internal energy of the gas change?

If the quantity of gas present is $2.00\times {10}^{-3}{\rm mol}$, find the molar specific heat of the gas at

(b) Constant pressure

(c) Constant volume.

A Carnot engine operates between $235{\rm{}^\circ\,\!C}$ and $115{\rm{}^\circ\,\!C}$, absorbing $6.30\times {10}^4{\rm J}$ per cycle at the higher temperature.

(a) What is the efficiency of the engine?

(b) How much work per cycle is this engine capable of performing?

A Carnot engine whose low temperature reservoir is at $17{\rm{}^\circ\,\!C}$ has an efficiency of $40\%$. By how much should the temperature of the high temperature reservoir be increased to increase the efficiency to $50\%$?

An ideal gas undergoes a reversible isothermal expansion at $77{\rm{}^\circ\,\!C}$. Increasing its volume from $1.30\ {\rm L}$ to $3.40\ {\rm L}$. The entropy change of the gas is $22\ {\rm J/K}$. How many moles of gas are present?

Four moles of an ideal gas undergo a reversible isothermal expansion from volume $V_1$ to volume $V_2=2V_1$ at temperature $T=400\ {\rm K}$. Find:

(a) The work done by the gas

(b) The entropy change of the gas

(c) If the expansion is reversible and adiabatic instead of isothermal, what is the entropy change of the gas?

A total work of $1590\ {\rm J}$ is applied to a moveable piston which causes the volume in a cylinder to contract. Assume this cylinder is filled with $84.6\ {\rm mol}$ of an ideal gas, which remains isobaric at a pressure of $8.6\ {\rm atm}$ as the piston moves. If the initial temperature of the gas is $12.6{\rm{}^\circ\,\!C}$, what is the final temperature of the gas (in ${\rm{}^\circ\,\!C}$) after the work is performed?

A quantity of neon gas (ideal and monoatomic) is compressed adiabatically in an insulated container from $0.200\ {{\rm m}}^{{\rm 3}}$ to $0.100\ {{\rm m}}^{{\rm 3}}$. The original pressure is $2.00\times {10}^5\ {\rm Pa}$.

(a) What is the final pressure?

(b) The final temperature of the neon gas is $305\ {\rm K}$. What is the initial temperature?

(c) What is the rms speed of the neon atoms in the final state? The mass of a single neon atom is $3.35\times {10}^{-26}\ {\rm k}{\rm g}$.

(d) How much does the internal energy of the gas change in this compression?

(e) How much work is done by the gas during this change?

A Carnot cycle is given as shown in the figure.AB and CD are isothermal with $T_h=500{\rm{}^\circ\,\!C}$ and $T_c=100{\rm{}^\circ\,\!C}$.BC and DA are adiabatic. The heat going into the cycle $Q_h$ is $2000{\rm J}$.

(a) Find the heat $Q_c$ leaving the cycle.

(b) Find the work done by the gas in a cycle.

(c) What is the efficiency of this cycle.

(a) We have an important relation for Carnot engine that relate temperatures and heat exchanges as
$\frac{\left|Q_c\right|}{Q_h}=\frac{T_c}{T_h}$
Use this relation to find the heat $Q_c$ released to the cold reservoir.
$\frac{\left|Q_c\right|}{Q_h}=\frac{T_c}{T_h}\ \Longrightarrow \ \left|Q_c\right|=Q_h\frac{T_c}{T_h}=2000\ \left(\frac{100+273}{500+273}\right)=965.1\ {\rm J\ }$
$\Rightarrow {{\rm Q}}_{{\rm c}}=-965.1\ {\rm J}$
We have inserted, by convention, a negative to indicate that the heat leaves the cycle.
(b) In the Carnot engines (cycles) from first law of thermodynamics, we have
$W_{net}+\left|Q_c\right|=Q_h$
$\ \Rightarrow \ W_{net}=2000-965.1=1034.9{\rm J}$
Where $W_{net}$ is the work done by the engine.
(c) The efficiency of an Carnot engine is given by
$e=1-\frac{T_c}{T_h}=1-\frac{373}{773}=0.517=51.7\%$

Consider the frictionless piston system below where gravity is pointing downwards and outside of the piston-gas system is vacuum.The piston mass is $M_p$ and its cross sectional area is $A$. The gas is ideal.

(a) Suppose the piston is initially held by hand at a certain initial height, and the initial pressure of the gas at that point is $P_1$. Subsequently, the piston is released and allowed to fall. Write down an inequality expression involving $M_p$ , $A$, $P_1$ , and $g$ which is necessary for the piston to fall. Express your answer in the form $P_1<\ \dots .$''.
(b) Suppose during the fall depicted in part (a), heat is drawn out of the gas such that the gas maintains a constant temperature as the piston falls. If the initial height of the piston is $h$, what is the final height of the piston? [Express your final answer in terms of $M_p$ , $A$, $P_1$ , $g$, and $h$.]

A projectile with mass ${\rm m}$ can escape'' from the surface of a planet if it is launched vertically upward with a kinetic energy greater than $mgR_p$ where ${\rm g}$ is the acceleration due to the gravity at the planet's surface and ${{\rm R}}_{{\rm p}}$ is the planet's radius. Ignore the air resistance.

(a) If the planet in the question is the earth, at what temperature does the average translational kinetic energy of the nitrogen molecule (molar mass $28.0\ {\rm g/mol}$) equal that required to escape the earth? What about the hydrogen molecule (molar mass $2.0\ {\rm g/mol}$), $g=9.8\ {\rm m/}{{\rm s}}^{{\rm 2}}$ , $R_E=6380\ {\rm km}$
(c) Repeat part (a) for the moon, which has $g=1.63\ {\rm m/}{{\rm s}}^{{\rm 2}}$ and $R_M=1740\ {\rm km}$. The moon and earth has similar surface temperatures, but the moon has no atmosphere and the earth has no $H_2$ in the atmosphere! Why is this the case?

(c) Repeat part (a) for the sun, which has $g=274.2\ {\rm m/}{{\rm s}}^{{\rm 2}}$ and $R_s=6.96\times {10}^5\ {\rm km}$. What does this say about escaping the sun, especially for those fleet-footed $H_2$ molecules. The surface temperature of the sun is 5800 K.

A heat engine takes $0.350\ {\rm moles}$ of a diatomic ideal gas around the cycle shown in the PV diagram. Process $1\to 2$ is at constant volume $V$ and process $2\to 3$ is adiabatic, and process $3\to 1$ is at a constant pressure P of $1.00\ {\rm atm}$. The value of $?$ for this gas is 1.4

(a) Find the pressure and volume at points 1,2 and 3.

(b) Calculate $Q\ ,\ W$ and $\Delta U$ for each of the three processes.

(c) Find the net work done by the gas in the cycle.

(d) Find the net heat flow into the engine in one cycle, use this result to calculate the thermal efficiency of this engine, and how it compares with the efficiency of a Carnot engine operating between the same $T_1$ and $T_2$?

You build a heat engine that takes ${\rm 1.00\ mol}$ of an ideal diatomic gas through the cycle shown. ($C_V=\frac{5}{2}R\ ,\ C_P=\frac{7}{2}R$)

(a) Show that the path or segment $ab$ is an isothermal compression.

(b) During which segment of the cycle is heat absorbed by the gas? During which segment of the cycle is heat rejected by the gas? How do you know this?

(c) Calculate the temperatures at points $a,b$ and $c$.

(d) Calculate the net heat exchanged with the surrounding heat reservoirs, and the net work done by the engine in one cycle, and calculate the thermal efficiency of the engine

(a) In an isothermal process, temperatures between two points are the same. So use the ideal gas law $PV=nRT$ to find $T_a$ and $T_b$.
$T_a=\frac{P_aV_a}{nR}=\frac{\left(2\times {10}^5\right)\left(0.01\right)}{\left(1\right)\left(8.314\right)}=240.55\ {\rm K}$
$T_b=\frac{P_bV_b}{nR}=\frac{\left(4\times {10}^5\right)\left(0.005\right)}{\left(1\right)\left(8.314\right)}=240.55\ {\rm K}$
Since $T_a=T_b$ so the path $b\to a$ is an isothermal process.
(b) Use the definitions of heat transferred $Q=nC_{V,P}\Delta T$ and work done on the system $W_{on}=-P\Delta V$ to find desired values. Path $ab$ is isothermal. Using first law of thermodynamics $\Delta U_{int}=W_{on}+Q_{in}$ we know that $\Delta U_{int}=0$ so $Q_{in}=-W_{on}=nRT{\ln \frac{V_b}{V_a}\ }\ \ ,\ \ V_a>V_b\ \Longrightarrow \ \ Q_{in}<0$ so heat is rejected.

In $bc$ process $P$ is constant, so $=nC_P\Delta T$ , use the ideal gas law to substitute $\frac{P\Delta V}{R}$ for $n\Delta T$ so  $Q=\frac{C_P}{R}P\Delta V$. Since $\Delta V=\left(V_c-V_a\right)>0$ so $Q>0$ and subsequently heat is absorbed.

In $ca$ process $V$ is constant (isochoric process), so $Q=nC_V\Delta T=\frac{C_V}{R}V\Delta P\ ,\ \Delta P<0\ \Longrightarrow Q<0\$so heat is rejected.

(c) $T_a=T_b=\frac{P_aV_a}{nR}=\frac{2\times {10}^3}{1\times 8.314}=240.55\ {\rm K}$
$T_c=\frac{P_cV_c}{nR}=\frac{4\times {10}^3}{1\times 8.314}=481.11\ {\rm K}$

(d) In part (c) we have find the temperatures of points, so
$Q_{bc}=nC_P\Delta {{\rm T}}_{{\rm bc}}{\rm =}\left({\rm 1.0}\right)\left(\frac{{\rm 7}}{{\rm 2}}\right)\left({\rm 8.134}\right)\left({\rm 481.11-240.55}\right){\rm =7.00\times }{{\rm 10}}^{{\rm 3}}\ {\rm J}$
$Q_{ca}=nC_V\Delta T_{ca}=\left(1.0\right)\left(\frac{5}{2}\right)\left(8.314\right)\left(240.55-481.11\right)=-5.00\times {10}^3{\rm J}$
$Q_{ab}=nRT{\ln \left(\frac{V_b}{V_a}\right)\ }=\left(1.0\right)\left(8.314\right)\left(240.55\right){\ln \left(\frac{0.005}{0.01}\right)\ }=-1.38\times {10}^3\ {\rm J}$
$\therefore Q_{total}=+620\ {\rm J}$
The path $ab$ is isothermal that is $\Delta U_{int}=\frac{3}{2}nR\Delta T={{\rm Q}}_{{\rm in}}{\rm +}{{\rm W}}_{{\rm on}}{\rm =0}$ ,thus the work done on the gas is ${\left(W_{on}\right)}_{ab}=-{\left(Q_{in}\right)}_{ab}=+1.38\times {10}^3\ {\rm J}$. By convention, the work done by the gas is ${(W_{by})}_{ab}=-1.38\times {10}^3{\rm J}$.
${\left(W_{on}\right)}_{bc}=-P\Delta V=-\left(4\times {10}^5\right)\left(0.01-0.005\right)=-2000\ {\rm J}$
${\left(W_{on}\right)}_{ca}=-P\Delta V=0$
$\therefore {\left(W_{on}\right)}_{net}=1380{\rm -}{\rm 2000=-620}\ {\rm J\ and\ }{\left(W_{by}\right)}_{net}{\rm =+620\ J}$
$e=\frac{{\left(W_{by}\right)}_{net}}{Q_H}=\frac{620}{7.01\times {10}^3}=8.84\ \%$

You have two identical containers, one containing gas A and the other gas B. the masses of these molecules are $m_A=3.3\times {10}^{-27}{\rm kg}$ and $m_B=5.34\times {10}^{-26}{\rm kg}$. Both gases are under the same pressure and are at temperature $T=10{\rm C{}^\circ\, =283\ K}$

(a) Which molecules, A or B have the greatest translational kinetic energy, and which molecules have the greatest ${\rm rms}$ speed $v_{rms}$? Recall the $v_{rms}={\left(\frac{3kT}{m}\right)}^{\frac{1}{2}}$

(b) Which container's temperature A or B should you raise so that both gasses will have the same rms speed $v_{rms}$.

(c) At what temperature will you accomplish this goal? Now which molecule A or B has the greatest translational energy?

(d) For part (c) what is the most probably speed $v_{mp}$ and what is the average speed $v_{ave}$ for the gas with the greatest energy. $v_{mp}={\left(\frac{2kT}{m}\right)}^{\frac{1}{2}}$, and $v_{ave}={\left(\frac{8kT}{pm}\right)}^{\frac{1}{2}}$.

The figure shows a $pV$ diagram for an ideal gas in which its absolute temperature at $b$ is one-fourth of its absolute temperature at $a$.

(a) What volume does the gas occupy at point $b$?

(b) How many joules of work was done by or on the gas in the process?

(c) By how much did the internal energy $U$ of the gas change in going from state $a$ to $b$?

(d) By how much did the heat content of the gas change as it went from state $a$ to $b$?

We are given $1/3$ of a mole of He gas, which is monoatomic. Calculate each of the following for the full path $a\to b\to c$.

(a) The work that is done.

(b) The change in internal energy.

A piston expands using an ideal gas under constant pressure of ${\rm 5\ atm}$ from ${\rm 0.40\ L}$ to ${\rm 0.80\ L}$. Heat then flows out of the piston under constant volume. The pressure is allowed to drop until the temperature is the same as the original temperature.
(a) Draw a PV-diagram for the process above
(b) Calculate the total work done by the process above
(c) Calculate the total heat flow into the system. Hint: First calculate the change in internal energy and then use the first law of thermodynamics.

An air mattress is inflated to a pressure of $110{\rm kPa}$(just a little above atmospheric pressure). When fully inflated, the mattress is $1.8\ {\rm m}$ long, $0.9\ {\rm m}$ wide and $15\ {\rm cm}$ thick. The mattress thickness gets decreased down to $13\ {\rm cm}$ when a person lies on it.

(a) If the length and width stay unchanged, what is the pressure of the air inside mattress when the person in lying on it?

(b) How many moles of air ($n$) are inside the mattress in part (a) if the temperature is $30{\rm{}^\circ\,\!C}$?

One mole of a monoatomic ideal gas is taken through the cycle shown on the P-V diagram. Initially, the gas occupies a volume $V_1=16\times {10}^{-3}{{\rm m}}^{{\rm 3}}$ at a pressure, $P_1=1.31\times {10}^5\ {\rm Pa}$. it is expanded at constant pressure$(1\to 2)$ until the volume occupied is $V_2=24\times {10}^{-3}{{\rm m}}^{{\rm 3}}$. The gas is then compressed adiabatically $(2\to 3)$ until it occupies the original volume. The pressure in the gas is brought back to its original value by isochoric process $(3\to 1)$. All processes are carried out slowly and reversibly.
(a) What is the temperature $T_2$ after the isobaric expansion $(1\to 2)$?

(b) What is the temperature $T_3$ after the adiabatic compression $(2\to 3)$?

(c) What is the heat absorbed by the gas $Q_{12}$ during the isobaric expansion $(1\to 2)$?

(d) Find the total work of the system?

A sample of Helium (ideal and monoatomic) is placed inside a piston. The walls of the piston are perfectly insulating and so no heat flows in or out the gas. Initially, the pressure is $1\ {\rm atm}$ and the volume is $1.00\times {10}^{-3}\ {{\rm m}}^{{\rm 3}}$. Now the pressure of the He is doubled, to $2\ {\rm atm}$. ($1\ {\rm atm=1.013\times }{{\rm 10}}^{{\rm 5}}\ {\rm Pa}$).

(a) What is the final volume of the Helium?

(b) The head of the piston has an area of $0.01{{\rm m}}^{{\rm 2}}$. What is the external force that must be exerted on the piston to maintain the Helium in its the compressed ($2\ {\rm atm}$) state?

The figure shows a $50{\rm kg}$ lead cylindrical piston which floats in equilibrium on $0.89\ {\rm mol}$ of compressed air at $30{\rm{}^\circ\,\!C}$. What is the height $h$ of the piston?

An expansion process on an ideal diatomic gas has a linear path between the initial and final states on a $PV$ diagram. The initial pressure is $300\ {\rm kPa}$, the initial volume is $0.090\ {{\rm m}}^{{\rm 3}}$, and the initial temperature is $390{\rm K}$. The final pressure is $80\ {\rm kPa}$ and the final temperature is $300\ {\rm K}$. How much work done by the gas in this process?

${\rm 4.8\ mol}$ of gas ${\rm \#1}$ initially has ${\rm 9000\ J\ }$of thermal energy. It interacts with ${\rm 3.4\ mol}$ of gas ${\rm \#2}$, which initially has ${\rm 5000\ J}$ of thermal energy. If both gases are monatomic, what is the change in thermal energy of gas ${\rm \#1}$?

$2.0\ {\rm moles}$ of Helium (monoatomic) at an initial temperature of $300\ {\rm K}$ come into contact with $4.0\ {\rm moles}$ Oxygen (diatomic) at an initial temperature of $600\ {\rm K}$. What is the final temperature?

The graph in the figure shows a cycle for a heat engine for which $Q_H=50\ {\rm J}$. What is the thermal efficiency of this engine?

The thermal efficiency of a heat engine is defined as the ratio of the magnitude of the work done by the engine to the magnitude of the heat absorbed from the high temperature reservoir
$e=\frac{\left|W\right|}{\left|Q_H\right|}$
Recall that the area under a PV diagram represents the work done by the gas so
\begin{align*}
W_{by}&=S_{triangle}\\
&=\frac{1}{2}\left(200\times {10}^{-6}-100\times {10}^{-6}\right)\left(400\times {10}^3-200\times {10}^3\right)\\
&=10\ {\rm J}
\end{align*}
So $e=\frac{\left|W\right|}{\left|Q_H\right|}=\frac{10}{50}=0.2=20\%$

An ideal monoatomic ideal gas cools from $455\,{\rm K}$ to $405\,{\rm K}$ at constant volume as $831\ {\rm J}$ of energy is removed from it. How many moles of gas are in the sample?

Assuming the radius of diatomic molecules is approximately $1.0\times {10}^{-10}{\rm m}$, for what pressure will the mean free path in room-temperature ($20{\rm{}^\circ\,\!C}$) nitrogen be $4.6{\rm m}$?

A sample of an ideal gas is slowly compressed to one-half its original volume with no change in temperature. What happens to the root-mean-square of the molecules in the sample?

A perfect Carnot engine operates between the temperatures of $300\,{\rm K}$ and $700\,{\rm K}$, drawing $60\,{\rm kJ}$ of heat from the $700\,{\rm K}$ reservoir in each cycle. How much heat is dumped into the $300\,{\rm K}$ reservoir in each cycle?

A Carnot refrigerator removes heat from the freezer at $-10{\rm{}^\circ\,\!C}$ and expels it into the room at $20{\rm{}^\circ\,\!C}$. You put a tray with $0.5\ {\rm kg}$ of water at $30{\rm{}^\circ\,\!C}$ into the freezer. ($c_{ice}=2.01\times {10}^3\frac{{\rm J}}{{\rm K.kg}}$, $c_w=4190\frac{{\rm J}}{{\rm K.kg}}$, $L_f=334\times {10}^3\frac{{\rm J}}{{\rm K}}$)
(a) Calculate the performance coefficient of this refrigerator.
(b) Calculate the heat energy extracted from this water when it is cooled and transformed to ice at $-10{\rm{}^\circ\,\!C}$.
(c) Calculate the work done by the compressor of the refrigerator in order to achieve this transformation (of part (b)).
(d) Calculate the heat that the refrigerator expels into the room while achieving this transformation.

An ideal gas initially at $300\ {\rm K}$ and occupying a volume of $20{\rm L}$ is adiabatically compressed. If its final temperature is $400{\rm K}$ and $\gamma=1.5$, what is its final volume?

In the $PV$ diagram on the right, $120\ {\rm J}$ of work was done by $0.12\ {\rm mol}$ of ideal gas during the adiabatic process $a\to b$. ($1\ {\rm atm=1.013\times }{{\rm 10}}^{{\rm 5}}\ {\rm Pa\ ,\ 1L=}{{\rm 10}}^{{\rm -}{\rm 3}}{{\rm m}}^{{\rm 3}}$)

(a) How much heat entered or left this gas from $a\$to $b$?

(b) By how much did the internal energy change?

(c) What is the temperature of the gas at $b$?

A metal rod has a thermal expansion coefficient of $6.44\times {10}^{-5}\ {{\rm{}^\circ\,\!C}}^{-1}$. Initially, the rod has a temperature of $55.6{}^\circ\, {\rm F}$ at some initial length. The rod is heated to $412\,{\rm {}^\circ\, F}$ which causes it length to change by $3.89\ {\rm mm}$. What must have been the initial length of the rod in ${\rm cm}$?

The change in length for a given change in temperature is $\Delta L=L_0a\Delta T$ where $a$ is the coefficient of linear expansion and  $T$ is absolute temperature (must be in Kelvins).
In this problem, the temperature has given in Fahrenheit so first must be converted to Celsius as $T\left({}^\circ\, F\right)=\frac{9}{5}T\left({\rm{}^\circ\,\!C}\right)+32$. Since an $1\ {\rm K}$ increase in temperature equals a $1{\rm{}^\circ\,\!C}$ that is $\Delta T_c=\Delta T_K$ use this note to convert the change in Fahrenheit to the change in Celsius as
$\Delta T_c=\frac{5}{9}{\Delta T}_F=\frac{5}{9}(412-55.6)=198{\rm{}^\circ\,\!C}$  since $\Delta T_c=\Delta T_K$ we have $\Delta L=L_0a\Delta T$
\begin{align*}
\Rightarrow \ \ L_0&=\frac{\Delta L}{a\Delta T}\\
&=\frac{3.89\ {\rm mm}}{\left({\rm 6.44\times }{{\rm 10}}^{{\rm -}{\rm 5}}\right)\left(198\right)}\\
&={\rm 305\ mm\times }\left(\frac{{\rm 1cm}}{{\rm 10mm}}\right)\\
&={\rm 30.5\ cm}
\end{align*}

A metal rod that is $92.2\ {\rm cm}$ long at $22.2{\rm{}^\circ\,\!C}$ is observed to be $94.6\ {\rm cm}$ long at $96.6{\rm{}^\circ\,\!C}$. What is the coefficient of linear expansion of this metal?

The ends of a cylindrical steel rod (with length $L$ and radius $R$) are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of $20\ {\rm cal/s}$. At what rate would a steel rod (with length $2L$ and radius $2R$) conduct heat between the same two temperatures?

Suppose that the insulating qualities of the wall of a house come mainly from a $0.100\ {\rm m}$ layer of brick with thermal conductivity $k_B=10.0\, {\rm J/(s.m.}{\rm{}^\circ\,\!C}{\rm )}$ and a $0.400\, {\rm m}$ layer of insulation with thermal conductivity $k_I=9.00\times {{\rm 10}}^{{\rm -}{\rm 2}}{\rm \ J/(s.m.}{\rm{}^\circ\,\!C}{\rm )}$. The temperature on the side of the brick is $T_B=10.0\,{\rm{}^\circ\,\!C}$ while the temperature on the side of the insulating material is $T_I=30.0{\rm{}^\circ\,\!C}$.
(a) What is the temperature at the interface brick/insulator?
(b) What is the total rate of heat loss$\ Q/t$ through such a wall, if the total area is $10.0\ {{\rm m}}^{{\rm 2}}$?
(c) A copper water pipe of length $1.00\ {\rm m}$ is moved from inside the house (at ${\rm 30.0}{\rm{}^\circ\,\!C}$) to the outside of the house (at $10.0\ {\rm{}^\circ\,\!C}$). What is the change in the length of this pipe after it adapts to the temperature? The coefficient of thermal linear expansion for copper is $\alpha=\ 1.70\times {10}^{-5}{\left({\rm{}^\circ\,\!C}\right)}^{-1}$ .

A glass windowpane is $1.7{\rm m}$ high, $1.4\ {\rm m}$ wide, and $2{\rm mm}$ thick. The temperature at the inner surface of the glass is $16{\rm{}^\circ\,\!C}$ and at the outer surface $4{\rm{}^\circ\,\!C}$. How much heat is lost each hour through the window under steady state conditions?

What is the net power that a person loses through radiation if her surface area is $1.20\ {{\rm m}}^{{\rm 2}}$, if her emissivity is $0.895$, if her skin temperature is $300{\rm K}$ and if she is in a room that is at a temperature of $17{\rm{}^\circ\,\!C}$?

A woman's body has a surface area of $1.4\ {{\rm m}}^{{\rm 2}}$ and her skin temperature is $27{\rm{}^\circ\,\!C}$. Her surroundings have a temperature of $17{\rm{}^\circ\,\!C}$. (recall that the emissivity can be taken to be 1 for human skin)
(a) Calculate the rate at which she emits heat radiation.
(b) Calculate the rate at which she absorbs heat radiation from the surroundings.
(c) What is the net rate at which she emits heat?

A $30000\ {\rm kg}$ subway train is initially at $20\ {\rm m/s}$. it slows to a stop at a station and stays there long enough for the brakes to cool. All the kinetic energy lost by the train is released as heat into the station, which is $10\ {\rm m}$ high, $20\ {\rm m}$ wide and $50\ {\rm m}$ long. Air has density of $1.2\ {\rm kg/}{{\rm m}}^{{\rm 3}}$ and a heat capacity of $1020\ {\rm J/(kg.K)}$.
(a) Calculate the mass of the air.
(b) Calculate the heat absorbed by the air.
(c) Calculate the change in the temperature of the air due to absorption of this heat.

The mechanical equivalent of heat is $\mathrm{1\ cal\ =\ 4.18\ J}$. The specific heat of water is $\mathrm{1\ cal/g.K}$, and its mass density is $\mathrm{1\ g/}{\mathrm{cm}}^{\mathrm{3}}$. An electric immersion water heater, rated at $\mathrm{400\ W}$, should heat a liter of water from $\mathrm{10{}^\circ C}$ to $\mathrm{30{}^\circ C}$ in about:
(a) $12\, \mathrm s$
(b) $50\, \mathrm s$
(c) $3.5\, \mathrm {min}$
(d) $15\, \mathrm {min}$
(e) $45\, \mathrm {min}$

First find the heat required  to change the temperature of the water from $10{}^\circ\!{C}$ to $30{}^\circ\!{C}$ and then using the definition of the power determine the time required as follows
Since there is no change in phase of water and only its temperature is changed so the heat required is calculated as $Q=mc\mathrm{\Delta }T$, where $c$ is the specific heat of the matter. Thus
$Q=\left(1000\ \mathrm{g}\right)\left(\mathrm{1}\frac{\mathrm{cal}}{\mathrm{g.K}}\right)\left(30-10\right)\left(\mathrm{K}\right)\mathrm{=20\times }{\mathrm{10}}^{\mathrm{3}}\mathrm{cal=20\ kcal}$
Note: the temperature difference of one Kelvin exactly coincides with one Celsius degree.
Power is defined as the ratio of the energy to the time. In the definition of power all quantities must be in SI units so we must convert Cal to Joules as
$20000\ \mathrm{cal\times 4.18\ J=83600\ J}$
$P=\frac{Q}{t}\Rightarrow t=\frac{Q}{P}=\frac{83600}{400}=209\mathrm{s=3.48\ min}$

Category : Thermodynamics

Most useful formula in Heat:

$T_C=\frac{5}{9}(T_F-32\,^{\circ}\mathrm{C} )$

Relation between Kelvins and Celsius:
$T_K=T_C+273.15$

Ideal gas law: $PV=nRT$

Ideal gas law between two states with fixed amount of gas:
$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

rms speed of molecules:
$v_{rms}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3RT}{M}}$
$m$ is the mass of the molecules and $M$ is the molar mass.

Mean free path $\lambda$:
$\lambda=\frac{1}{\sqrt{2}n_V\pi d^2}$
$d$= molecule’s diameter, $n_V$= number of molecules per unit volume

Heat capacities:
$C_V=\frac{3}{2}R \ \ (monatomic \ gas)$
$C_V=\frac{5}{2}R \ \ (diatomic \ gas)$

Relation between $C_V$ and $C_P$:
$C_P=C_V+R$

Thermal expansion:
$\Delta L=\alpha L_0 \Delta T$
$\Delta V=\beta V_0 \Delta V$

Required heat to a temperature change $\Delta T$, without phase change
$Q=mc\Delta T$
Required heat to change in phase of a matter:
$Q=\pm mL$
$L$ is the heat of fusion $L_f$, vaporization $L_V$

Number Of Questions : 62