Heat and Thermodynamics

(b) $\frac{R}{\gamma-1}$

Mayer's relation: $C_p-C_v=R$, and $C_p=\gamma C_v$. So $(\gamma-1)C_v=R$, giving $C_v=\frac{R}{\gamma-1}$.

(b) Wien's displacement law

As the temperature rises, the wavelength of maximum emission shifts to shorter wavelengths: $\lambda_mT=\text{constant}$. That is why the colour moves from dull red to yellow to white. Stefan's law only gives the total power radiated, not the colour.

(d) $\frac32$

For an adiabatic process $P^{1-\gamma}T^{\gamma}=\text{const}$, so $P\propto T^{\gamma/(\gamma-1)}$. Given $P\propto T^3$: $\frac{\gamma}{\gamma-1}=3$, so $\gamma=3\gamma-3$ and $\gamma=\frac32$.

(a) $\frac38N_ak_B(T_2-T_1)$

Helium is monatomic with molar mass 4 g, so 1 g is $\frac14$ mole, and $C_v=\frac32R$. $Q=nC_v\Delta T=\frac14\cdot\frac32R\,\Delta T=\frac38R\Delta T$, and $R=N_ak_B$.

(b) $P/64$

Isothermal ($V\to2V$): $P\cdot V=P_1\cdot2V$, so $P_1=\frac{P}{2}$. Adiabatic ($2V\to16V$): $P_2=P_1\left(\frac{2V}{16V}\right)^{5/3}=\frac{P}{2}\times\left(\frac18\right)^{5/3}=\frac{P}{2}\times\frac{1}{32}=\frac{P}{64}$. Options (a), (c) and (d) are all larger than $P$, which is impossible for an expansion.

(d) $45^\circ$C

By Newton's law, $\frac{\Delta T}{t}=k(T_{avg}-T_0)$. First interval: $\frac{10}{5}=k(65-T_0)$. Second interval: $\frac{6}{5}=k(57-T_0)$. Dividing: $\frac{10}{6}=\frac{65-T_0}{57-T_0}$, so $570-10T_0=390-6T_0$, giving $T_0=45^\circ$C. The surroundings must be cooler than $54^\circ$C, as the water is still cooling.

(c) 22.5 g

Heat lost by $m$ g of steam condensing and cooling to $80^\circ$C equals heat gained by $20$ g of water: $m(540)+m(1)(100-80)=20\times1\times(80-10)$. $560m=1400$, so $m=2.5$ g of steam condensed. Total water present $=20+2.5=22.5$ g. Option (a) wrongly adds an extra amount of condensed mass.

(a) $r^2$

Mean free path $\lambda=\frac{1}{\sqrt2\,\pi d^2 n}$, where $d=2r$ is the molecular diameter. So $\lambda\propto\frac{1}{r^2}$, i.e. it is inversely proportional to $r^2$. Collision cross-section depends on the area $\pi d^2$, which explains the square, not the first or third power.

(c) Adiabatic

Compare the $P$-$V$ curves from the same initial state to half the volume: the adiabat is steepest, so it ends at the highest pressure and work done on the gas (area under the curve) is greatest. The order is: isobaric $<$ isothermal $<$ adiabatic, as $PV^\gamma$ rises faster than $PV$. Isochoric involves no volume change, so no work, which rules out (a).

(c) $8.0\ \mathrm{J\,K^{-1}\,mol^{-1}}$

4.0 g occupying 22.4 L at NTP is 1 mol, so $M = 4\times10^{-3}$ kg/mol. From $v=\sqrt{\gamma RT/M}$ with $T=273$ K: $\gamma = \dfrac{v^2M}{RT} = \dfrac{(952)^2(4\times10^{-3})}{8.3\times273}\approx1.6$. Then $C_p=\gamma C_v = 1.6\times5.0 = 8.0\ \mathrm{J\,K^{-1}\,mol^{-1}}$.

(c) $31^\circ$C

For a refrigerator, COP $=\dfrac{T_2}{T_1-T_2}$ with $T_2=-20^\circ\text{C}=253$ K. So $5=\dfrac{253}{T_1-253}$, giving $T_1-253=50.6$ and $T_1\approx303.6$ K, which is about $31^\circ$C.

(d) 0.020

Density varies inversely with volume, so for small changes $\dfrac{\Delta\rho}{\rho}\approx-\dfrac{\Delta V}{V}=-\gamma\Delta T$. Magnitude: $5\times10^{-4}\times40=0.020$ (a decrease).

(a) $\alpha_1 l_1=\alpha_2 l_2$

After a temperature rise $\Delta T$, the lengths are $l_1(1+\alpha_1\Delta T)$ and $l_2(1+\alpha_2\Delta T)$. Their difference stays constant only if the expansions are equal: $l_1\alpha_1\Delta T=l_2\alpha_2\Delta T$, giving $\alpha_1l_1=\alpha_2l_2$.

(d) 236.5 W

A refrigerator has $\text{COP}=\dfrac{T_2}{T_1-T_2}=\dfrac{277}{303-277}=\dfrac{277}{26}\approx10.65$. Heat removed per second is $Q_2=600\times4.2=2520$ J, so the power is $W=\dfrac{Q_2}{\text{COP}}=\dfrac{2520}{10.65}\approx236.5$ W.

(c) Compressing the gas through adiabatic process will require more work to be done.

On a $P$–$V$ diagram the adiabat is steeper than the isotherm, since $PV^\gamma=\text{const}$ with $\gamma>1$. For the same volume reduction, the adiabatic pressure rises more (the gas also heats up), so the area under it, which is the work, is larger. This holds for every gas, so atomicity (option a) does not matter.

(c) $\dfrac{400}{\sqrt3}$

The r.m.s. speed depends only on temperature, $v_{rms}\propto\sqrt T$, not on pressure. So $v=200\sqrt{\dfrac{400}{300}}=200\cdot\dfrac2{\sqrt3}=\dfrac{400}{\sqrt3}\ \mathrm{m/s}$. The pressure change is a distractor.

(d) 136 km

The energy lost in the fall is $mgh$, and only one quarter of it melts the ice: $\frac14mgh=mL$. So $h=\dfrac{4L}{g}=\dfrac{4\times3.4\times10^5}{10}=1.36\times10^5$ m $=136$ km. Mass cancels.

(c) 1800

Power radiated $P=\sigma A T^4\propto r^2T^4$.

Halving $r$ gives $\tfrac14$, doubling $T$ gives $16$, so $P'=450\times\tfrac14\times16=1800$ W.

(c) $11RT$

Neglecting vibration, $\mathrm{O_2}$ (diatomic) has $f=5$ and Ar (monatomic) has $f=3$. $U=\sum n\tfrac f2RT$.

$U=2\times\tfrac52RT+4\times\tfrac32RT=5RT+6RT=11RT$.

(a) 90 J

As a heat engine $\eta=1-\tfrac{T_2}{T_1}=\tfrac1{10}$, so $T_2/T_1=0.9$.

As a refrigerator, $\mathrm{COP}=\dfrac{T_2}{T_1-T_2}=\dfrac{0.9}{0.1}=9$. Heat absorbed from the cold reservoir is $Q_2=\mathrm{COP}\times W=9\times10=90$ J.

(c) $\dfrac{256}{81}$

Wien's law: $\lambda_mT=$ constant. With $\lambda\to\tfrac34\lambda_0$, $T\to\tfrac43T$.

Power $\propto T^4$, so $n=(4/3)^4=256/81$. Using $(3/4)^4$ would invert the relation.

(d) 208.7 J

Heat supplied $Q = 54 \times 4.18 \approx 225.7$ J. Work done by the steam against the atmosphere is $P\Delta V = 1.013\times10^5 \times 167.1\times10^{-6} \approx 16.9$ J (the water volume is negligible). First law: $\Delta U = Q - W \approx 225.7 - 16.9 \approx 208.7$ J. Option (b) wrongly subtracts too much work.

(b) 26.8%

The ideal (Carnot) efficiency is $\eta = 1 - T_2/T_1$ with temperatures in kelvin: $T_2 = 273$ K, $T_1 = 373$ K. $\eta = 1 - 273/373 \approx 0.268 = 26.8\%$. Using Celsius values would give a wrong answer.

(d) $8.360\times10^{4}$ K

Escape speed $v_e = 11.2$ km/s, and $v_{rms} = \sqrt{3k_BT/m}$. Setting these equal: $T = \dfrac{m v_e^2}{3k_B}$. $T = \dfrac{2.76\times10^{-26}\times(1.12\times10^{4})^2}{3\times1.38\times10^{-23}} \approx 8.36\times10^{4}$ K. Omitting the factor 3 gives option (a)-style values.

(b) increase in its kinetic energy

Mean kinetic energy of the molecules is $\tfrac32 k_BT$, so raising the temperature increases it. The mass of the gas is unchanged, the pressure in a closed container rises (not falls), and the average intermolecular distance stays the same in a fixed volume.

(d) 68 cm

For equal increase in length at any temperature rise, $L_{Cu}\alpha_{Cu}\Delta T=L_{Al}\alpha_{Al}\Delta T$. $L_{Al}=88\times\dfrac{1.7}{2.2}=68$ cm. The longer rod must have the smaller $\alpha$, so 113.9 cm (the inverted ratio) is wrong.

(d) $\mathrm{W\,m^{-1}\,K^{-1}}$

From $H=\dfrac{KA\,\Delta T}{L}$, $K=\dfrac{HL}{A\,\Delta T}$ with $H$ in watt (J/s). Units: $\dfrac{\mathrm{W\cdot m}}{\mathrm{m^2\cdot K}}=\mathrm{W\,m^{-1}\,K^{-1}}$, equivalent to $\mathrm{J\,s^{-1}\,m^{-1}\,K^{-1}}$. Option (b) lacks the per-second.

(b) Adiabatic

In an adiabatic process $Q=0$: no heat enters or leaves the system, so $\Delta U=-W$. Isothermal needs heat exchange to keep $T$ constant, isobaric and isochoric processes exchange heat as $T$ changes.

(d) $\dfrac{3}{2}k_BT$

A monatomic gas has 3 translational degrees of freedom, each with $\tfrac12k_BT$ by equipartition. So the average energy per molecule is $\tfrac32k_BT$. $\tfrac52k_BT$ is for a diatomic gas (rotation included).

(c) $\dfrac{27}{8}$

Heat needed for a rise of 1 K is $Q=mc\Delta T\propto m$. Both spheres are copper, so $m\propto$ volume $\propto r^3$. $\dfrac{Q_1}{Q_2}=\left(\dfrac{r_1}{r_2}\right)^3=(1.5)^3=\dfrac{27}{8}$. The ratio $\tfrac94$ would be for surface area.

(d) $\frac{13}{5}t$

By Newton's law of cooling, rate $\propto$ (mean temperature $-$ room temperature). First case: $\frac{10}{t}=k(85-20)=65k$. Second case: $\frac{20}{t'}=k(70-20)=50k$. Dividing, $t'=\frac{20\times65}{10\times50}t=\frac{13}{5}t$.

(a) A-Q, B-P, C-S, D-R

$v_{rms}=\sqrt{\frac{3RT}{M}}$ (Q); pressure $P=\frac{1}{3}nm\bar{v^2}$ (P); average kinetic energy per molecule $=\frac{3}{2}k_BT$ (S); internal energy of 1 mole of a diatomic gas (5 degrees of freedom) $=\frac{5}{2}RT$ (R).

(d) $5.6\ \mathrm{m^3}$

4.5 kg of water is $4500/18 = 250$ mol. With no intermolecular forces it behaves as an ideal gas, and 1 mol occupies 22.4 L at STP: $250\times22.4 = 5600$ L $= 5.6\ \mathrm{m^3}$.

(a) $27^\circ$C

$\eta = 1 - T_2/T_1$ with $T_1 = 327+273 = 600$ K. $0.5 = 1 - T_2/600$, so $T_2 = 300$ K $= 27^\circ$C. Temperatures must be in kelvin.

(b) $3295^\circ$C

$T_1 = -50+273 = 223$ K. "Increased by 3 times" means the new rms speed is $v+3v=4v$. Since $v_{rms}\propto\sqrt{T}$, $T_2 = 16\times223 = 3568$ K $=3295^\circ$C. Treating it as a 3-fold speed gives only $9T_1\approx1734^\circ$C, which matches no option; 669°C and 223 K are not consistent with either reading.

(a) $50\times10^{3}$ N

The bar is not allowed to expand, so thermal strain $\alpha\Delta T$ is fully converted into compressive stress: $F=YA\alpha\Delta T$. $F=0.5\times10^{11}\times10^{-3}\times10^{-5}\times100=5\times10^{4}$ N $=50\times10^3$ N. The length (1 m) cancels out and does not affect the force.

(a) $\frac{5}{3}$

Thermal resistance of each rod is proportional to $\frac1k$: $\frac12,\ 1,\ \frac12$ (in common units), total $2$. The temperature drop is $3T-T=2T$, so the heat current is proportional to $\frac{2T}{2}=T$. $T_1=3T-T\cdot\frac12=\frac52T$ and $T_2=T+T\cdot\frac12=\frac32T$. $\frac{T_1}{T_2}=\frac53$.

(d) 1.6 atm

The partition is insulating and the total volume is fixed, so no work is done and internal energy is conserved: $n_1T_1+n_2T_2=(n_1+n_2)T$. From $pV=nRT$, $n_1RT_1=2$ and $n_2RT_2=6$ (atm L). So $9RT=8$. $p=\frac{(n_1+n_2)RT}{V}=\frac{8}{5}=1.6$ atm. Simply adding $pV$ values (10/5 = 2 atm) is wrong because the initial temperatures differ.

(a) 0.116 kg

Final absolute pressure $=12$ atm $=12.12\times10^5$ Pa (gauge 11 plus 1 atm). $RT=\frac{100}{12}\times300=2500$ J/mol. $n_f=\frac{PV}{RT}=\frac{12.12\times10^5\times0.030}{2500}=14.54$ mol. Withdrawn $=18.20-14.54=3.66$ mol. Mass $=3.66\times32\approx117$ g $\approx0.116$ kg. Using the gauge pressure directly would give a wrong answer.

(d) $\frac{3}{4}$

At constant pressure $Q=\Delta U+W$. The heat and $\Delta U$ are equal, so the work done $W=P\Delta V$ is equal for both gases. $P\,\pi r_A^2(16)=P\,\pi r_B^2(9)$, so $\frac{r_A^2}{r_B^2}=\frac{9}{16}$. $\frac{r_A}{r_B}=\frac34$.