Thermodynamics

Answer: (b)

1. Boyle's law at constant $T$: $PV = \text{constant}$, so $P' = \dfrac{PV}{V'}$.
2. $V' = 0.96V \Rightarrow P' = \dfrac{P}{0.96} = 1.0417P$
3. Increase $\approx 4.17\%$, i.e. about $4.16\%$.
4. So the answer is (b) $4.16\%$.

Answer: (b)

1. Gas equation: $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$, with $T_1 = 300$ K, $T_2 = 270$ K.
2. $V_2 = V_1 \cdot \dfrac{P_1}{P_2} \cdot \dfrac{T_2}{T_1} = 500 \times \dfrac{1}{0.5} \times \dfrac{270}{300}$
3. $V_2 = 500 \times 2 \times 0.9 = 900$ m$^3$
4. So the answer is (b) $900$ m$^3$.

Answer: (d)

1. Boyle's law: $V \propto \dfrac{1}{P}$ at constant $T$.
2. $P' = 0.8P \Rightarrow V' = \dfrac{V}{0.8} = 1.25V$
3. So the answer is (d): volume increases by $25\%$.

Answer: (a)

1. Charles's law: $V \propto T$ at constant $P$; $T_1 = 300$ K, $T_2 = 900$ K.
2. $V_2 = V_1 \dfrac{T_2}{T_1} = 4 \times \dfrac{900}{300} = 12$ m$^3$
3. So the answer is (a) $12$ m$^3$.

Answer: 3730

1. Closed cylinder (constant $V$): $P \propto T$ (in kelvin).
2. $T_1 = 50^\circ$C $= 323$ K; doubled temperature $100^\circ$C $= 373$ K.
3. $P_2 = 3230 \times \dfrac{373}{323} = 3730$ Pa
4. Answer: $3730$ Pa

Answer: (c)

1. Fixed volume: $P = \dfrac{nRT}{V} \Rightarrow \dfrac{P_2}{P_1} = \dfrac{n_2}{n_1}\cdot\dfrac{T_2}{T_1}$.
2. $n_2 = \dfrac{n_1}{2}$, $T_1 = 300$ K, $T_2 = 350$ K.
3. $P_2 = 20 \times \dfrac{1}{2} \times \dfrac{350}{300} = 11.67 \approx 12$ atm
4. So the answer is (c) $12$ atm.

Answer: (d)

1. $PV = nRT$ with $n = \dfrac{m}{M} = \dfrac{7}{32}$.
2. So the answer is (d) $PV = \left(\dfrac{7}{32}\right)RT$.

Answer: (d)

1. $PV = \dfrac{m}{M}RT$ with $V$ fixed $\Rightarrow m \propto \dfrac{P}{T}$.
2. $m_2 = 6 \times \dfrac{P/2}{P} \times \dfrac{500}{300} = 6 \times \dfrac{1}{2} \times \dfrac{5}{3} = 5$ g
3. Leaked mass $= 6 - 5 = 1$ g
4. So the answer is (d) $1$ gram.

Answer: 481

1. Volume constant, so $Q = nC_V\Delta T$ with $C_V = C_p - R = \frac{5}{2}R$.
2. $\Delta T = 50 - 20 = 30\text{ K}$.
3. $n = \frac{300}{\frac{5}{2} \times 8.314 \times 30} = \frac{300}{623.6} \approx 0.481$ mol.
4. Expressed as asked (in units of $10^{-3}$), the answer is 481.

Answer: 7

1. Same gas: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.
2. $T_1 = 300\text{ K}$, $T_2 = 350\text{ K}$.
3. $P_2 = P_1\frac{V_1}{V_2}\cdot\frac{T_2}{T_1} = 2 \times \frac{60}{20} \times \frac{350}{300} = 7$ atm.
4. Final pressure = 7 atm.

Answer: (d)

1. $P = \frac{\rho RT}{M} \Rightarrow \frac{\rho}{P} = \frac{M}{RT} \propto \frac{1}{T}$.
2. $T_1 = 283\text{ K}$, $T_2 = 383\text{ K}$.
3. New ratio $= x\cdot\frac{283}{383}$.
4. So the answer is (d) $\frac{283x}{383}$.

Answer: (b)

1. Number of molecules $N = \frac{PV}{kT} \propto \frac{PV}{T}$.
2. $N_A \propto \frac{PV}{T}$, $N_B \propto \frac{2P\cdot\frac{V}{4}}{2T} = \frac{1}{4}\frac{PV}{T}$.
3. $N_B : N_A = 1 : 4$.
4. So the answer is (b) $1 : 4$.

Answer: (b)

1. Moles at start $n_1 = \frac{PV}{RT}$; after leakage $n_2 = \frac{P'V}{RT}$ ($V$, $T$ same).
2. Leaked moles $= n_1 - n_2 = \frac{V}{RT}(P - P')$.
3. So the answer is (b) $\frac{V}{RT}(P - P')$.

Answer: (d)

1. Pressure at bottom $P_1 = P_0 + \rho g h = 10^5 + 10^3 \times 10 \times 5 = 1.5 \times 10^5\text{ Pa}$; at surface $P_2 = 10^5\text{ Pa}$.
2. $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ with $T_1 = 290\text{ K}$, $T_2 = 300\text{ K}$.
3. $V_2 = 2.9 \times 1.5 \times \frac{300}{290} = 4.5\text{ cm}^3$.
4. So the answer is (d) 4.5.

Answer: (c)

1. Total moles $n = \frac{PV}{RT} = \frac{10^5 \times 8.31 \times 10^{-3}}{8.31 \times 300} = \frac{1}{3} \approx 0.333$ mol.
2. Let $x$ = moles of CO₂, $y$ = moles of O₂: $x + y = \frac{1}{3}$ and $44x + 32y = 13.2$.
3. $44x + 32\left(\frac{1}{3} - x\right) = 13.2 \Rightarrow 12x = 2.53 \Rightarrow x \approx 0.21$, $y \approx 0.12$.
4. So the answer is (c) 0.21 and 0.12.

Answer: (a)

1. Total moles are conserved and the final pressure $P$ is common: $\frac{90V}{400} + \frac{90V}{400} = \frac{PV}{400} + \frac{PV}{500}$.
2. $0.45 = P(0.0025 + 0.0020) = 0.0045P$.
3. $P = 100$ kPa.
4. So the answer is (a) 100.

Answer: (c)

1. Gas equation: $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$, with $T_1 = 300$ K, $T_2 = 270$ K.
2. $V_2 = 500 \times \dfrac{1}{0.5} \times \dfrac{270}{300} = 900$ m$^3$
3. So the answer is (c) $900$ m$^3$.

Answer: (c)

1. Boyle's law: $P \propto \dfrac{1}{V}$ at constant $T$.
2. $V' = 0.93V \Rightarrow P' = \dfrac{P}{0.93} = 1.0753P$
3. Increase $\approx 7.5\%$.
4. So the answer is (c) $7.52\%$.

Answer: (c)

1. Boyle's law: $P \propto \dfrac{1}{V}$ at constant $T$.
2. $V' = 0.95V \Rightarrow P' = \dfrac{P}{0.95} = 1.0526P$
3. So the answer is (c) $5.26\%$.

Answer: (d)

1. Volume of tyre fixed, so $P \propto T$; $T_1 = 300$ K, $T_2 = 310$ K.
2. $P_2 = 270 \times \dfrac{310}{300} = 279$ kPa
3. So the answer is (d) 279 kPa.

Answer: (d)

1. Average translational KE per molecule $= \dfrac{3}{2}k_BT$, independent of molar mass.
2. Both gases are at 300 K, so the KEs are equal.
3. So the answer is (d) 1 : 1.

Answer: (d)

1. Closed vessel (constant $V$): $P \propto T \Rightarrow \dfrac{\Delta P}{P} = \dfrac{\Delta T}{T}$.
2. $0.004 = \dfrac{1}{T} \Rightarrow T = 250$ K
3. So the answer is (d) 250 K.

Answer: 327

1. Fixed volume: $P \propto T$, so doubling $P$ needs doubling $T$ in kelvin.
2. $T_1 = 300$ K $\Rightarrow T_2 = 600$ K $= 327^\circ$C
3. Answer: $327^\circ$C

Answer: (c)

1. Closed vessel: $P \propto T \Rightarrow \dfrac{\Delta P}{P} = \dfrac{\Delta T}{T}$.
2. $0.023 = \dfrac{4}{T} \Rightarrow T = \dfrac{4}{0.023} \approx 174$ K
3. No option matches exactly; with $2.5\%$ the result is $\dfrac{4}{0.025} = 160$ K.
4. So the intended answer is (c) 160 K.

Answer: (a)

1. Absolute pressure = gauge + 1 atm $= 12$ atm $= 12 \times 1.01 \times 10^5 = 1.212 \times 10^6$ Pa.
2. Moles left: $n = \dfrac{PV}{RT} = \dfrac{1.212 \times 10^6 \times 0.030}{\frac{100}{12} \times 300} = \dfrac{36360}{2500} = 14.54$ mol
3. Moles withdrawn $= 18.20 - 14.54 = 3.66$ mol
4. Mass $= 3.66 \times 32 \approx 117$ g $\approx 0.116$ kg
5. So the answer is (a) 0.116 kg.

Answer: (b)

1. $N \propto \frac{PV}{T}$.
2. $N_A \propto \frac{PV}{T}$, $N_B \propto \frac{2P\cdot\frac{V}{4}}{\frac{T}{4}} = \frac{2PV}{T}$.
3. $N_A : N_B = 1 : 2$.
4. So the answer is (b) $1 : 2$.

Answer: (d)

1. $\rho = \frac{PM}{RT} \Rightarrow \rho \propto \frac{P}{T}$.
2. $\frac{\rho}{\rho_0} = \frac{3P_0}{P_0}\cdot\frac{T_0}{2T_0} = \frac{3}{2}$.
3. So the answer is (d) $\frac{3}{2}\rho_0$.

Answer: (b)

1. Moles before $= \frac{PV}{RT}$, after $= \frac{P'V}{RT}$ (same $V$ and $T$).
2. Leaked $= \frac{V}{RT}(P - P')$.
3. So the answer is (b) $\frac{V}{RT}(P - P')$.

Answer: (d)

1. For an ideal gas $PV$ is proportional to internal energy, which is conserved when the partition is removed.
2. $p(V_1 + V_2) = p_1V_1 + p_2V_2$.
3. $p = \frac{1 \times 2 + 2 \times 3}{2 + 3} = \frac{8}{5} = 1.6$ atm.
4. So the answer is (d) 1.6 atm.

Answer: (d)

1. Total moles are conserved: $\frac{P_1(2V)}{R \cdot 1000} + \frac{P_2V}{R \cdot 500} = \frac{P(3V)}{R \cdot 600}$.
2. $\frac{8 \times 2}{1000} + \frac{7}{500} = \frac{3P}{600}$.
3. $0.016 + 0.014 = 0.030 = \frac{P}{200} \Rightarrow P = 6$ kPa.
4. So the answer is (d) 6.

Answer: (c)

1. Boyle's law: $V \propto \dfrac{1}{P}$ at constant $T$.
2. $P' = 0.8P \Rightarrow V' = \dfrac{V}{0.8} = 1.25V$
3. So the answer is (c): volume increases by $25\%$.

Answer: (a)

1. Closed vessel: $P \propto T \Rightarrow \dfrac{\Delta P}{P} = \dfrac{\Delta T}{T}$.
2. $T = 100^\circ$C $= 373$ K.
3. $\Delta T = 0.04 \times 373 \approx 14.9^\circ$C
4. This matches none of the options; key gives (a) $2^\circ$C.

Answer: (c)

1. $PV = \dfrac{m}{M}RT$; same $m$, $P$, $V$ $\Rightarrow T \propto M$.
2. $\dfrac{T_{O_2}}{T_{H_2}} = \dfrac{32}{2} = 16$
3. So the answer is (c) $16 : 1$.

Answer: (d)

1. Charles's law: $V = \dfrac{nR}{P}T \Rightarrow T = \left(\dfrac{P}{nR}\right)V$, so slope of the $T$–$V$ line $\propto P$.
2. Dotted line for $P_1$ is steepest, then $P_2$, then $P_3$.
3. So the answer is (d) $P_1 > P_2 > P_3$.

Answer: (a)

1. $P = \frac{\rho R T}{M}$, so for a fixed density the $P$-$T$ graph is a straight line of slope $\frac{\rho R}{M}$.
2. Slope $\propto \rho$: the steepest line has the largest density.
3. From the figure, slope of $\rho_1$ > slope of $\rho_2$ > slope of $\rho_3$, so $\rho_1 > \rho_2 > \rho_3$.
4. So the answer is (a) $\rho_1 > \rho_2$.

Answer: (a)

1. $P = nkT$, where $n$ is the number of molecules per m³.
2. $T = \frac{P}{nk} = \frac{1.38 \times 10^5}{2.0 \times 10^{25} \times 1.38 \times 10^{-23}}$ (taking 1 atm $\approx 10^5$ Pa).
3. $T = \frac{10^5}{200} = 500\text{ K}$.
4. So the answer is (a) 500 K.

Answer: (c)

1. $\rho = \frac{PM}{RT} \Rightarrow M \propto \frac{\rho}{P}$ at the same $T$.
2. $\frac{M_A}{M_B} = \frac{\rho_A}{\rho_B}\cdot\frac{P_B}{P_A} = 1.5 \times \frac{1}{2} = \frac{3}{4}$.
3. So the answer is (c) $3 : 4$.

Answer: (d)

1. Same $V$ and $T$, so $P \propto n$.
2. $n_A = \frac{1}{2}$ (H₂), $n_B = \frac{1}{32}$ (O₂).
3. $\frac{P_A}{P_B} = \frac{1/2}{1/32} = 16$.
4. So the answer is (d) 16.

Answer: (d)

1. Boyle's law: $V \propto \dfrac{1}{P}$ at constant $T$.
2. $P' = 1.05P \Rightarrow V' = \dfrac{V}{1.05} = 0.9524V$
3. Decrease $\approx 4.76\%$, nearest option $4.70\%$.
4. So the answer is (d) $4.70\%$.

Answer: (b)

1. Charles's law: at constant $P$, $V = \left(\dfrac{nR}{P}\right)T$, so $V \propto T$.
2. The $V$–$T$ graph is a straight line of positive slope that passes through the origin when extended.
3. Graph Q shows this rising straight line; P falls, R is not a single line, S is constant.
4. So the answer is (b) Q.

Answer: (a)

1. Unit of $PV$ = $\dfrac{\text{N}}{\text{m}^2} \times \text{m}^3 = \text{N m} = \text{J}$.
2. So the answer is (a) energy.

Answer: (d)

1. Volume fixed, so $P \propto T$; $T_1 = 300$ K, $T_2 = 309$ K.
2. $P_2 = 270 \times \dfrac{309}{300} = 278.1$ kPa
3. So the answer is (d) 278 kPa.

Answer: (a)

1. At constant volume, $P \propto T$ (absolute), so each $P$–$t$ line reaches $P = 0$ at absolute zero.
2. All lines meet the temperature axis at K, so K is absolute zero.
3. So the answer is (a) $-273^\circ$C.

Answer: (d)

1. $PV = \dfrac{m}{M}RT$; same $m$, $P$, $V$ $\Rightarrow T \propto M$.
2. $\dfrac{T_{O_2}}{T_{H_2}} = \dfrac{32}{2} = 16$
3. So the answer is (d) $16 : 1$.

Answer: (d)

1. Dalton's law: total moles $n = 0.2 + 0.3 = 0.5$; $V = 200$ cm$^3 = 2 \times 10^{-4}$ m$^3$.
2. $P = \dfrac{nRT}{V} = \dfrac{0.5 \times 8.3 \times 200}{2 \times 10^{-4}} = \dfrac{830}{2 \times 10^{-4}}$
3. $P = 4.15 \times 10^6$ Pa
4. So the answer is (d) $4.15 \times 10^{6}$ Pa.

Answer: (d)

1. For an ideal gas $\gamma = 1 + \frac{2}{f}$ depends only on the degrees of freedom $f$.
2. It does not change with temperature, so $\gamma \propto T^0$.
3. So the answer is (d) $\gamma \propto T^{0}$.

Answer: (d)

1. From $PV = nRT$: $P \propto \frac{T}{V}$.
2. $PT^2 =$ const $\Rightarrow \frac{T}{V}T^2 =$ const $\Rightarrow V \propto T^3$.
3. Volume expansion coefficient $\gamma_V = \frac{1}{V}\frac{dV}{dT} = \frac{3}{T}$.
4. So the answer is (d) $\frac{3}{T}$.

Answer: (a)

1. $\rho = \frac{PM}{RT} \Rightarrow \rho \propto \frac{P}{T}$.
2. $\frac{\rho}{\rho_0} = \frac{3P_0}{P_0}\cdot\frac{T_0}{2T_0} = \frac{3}{2}$.
3. So the answer is (a) $\frac{3}{2}\rho_0$.