JEE Physics Thermodynamics MCQs Set 13

Answer: (b) 32 °C

Explanation:
1. Moles: oxygen = 16/32 = 0.5 mol, nitrogen = 14/28 = 0.5 mol, both with \( C_v=\tfrac52R \).
2. Energy conservation: \( n_1C_v(T-310)+n_2C_v(T-300)=0 \).
3. With equal moles, \( T=\dfrac{310+300}{2}=305 \) K = 32 °C.

Teacher's Note:
1. Because both gases are diatomic, the common factor \( C_v \) cancels out.
2. Convert to kelvin before averaging, then convert the answer back.

Answer: (a) 196 °C

Explanation:
1. n = 0.040/4 = 0.01 mol and \( \Delta U=\tfrac32nR\Delta T \).
2. \( \Delta T=\dfrac{12}{1.5\times0.01\times25/3}=96 \) K.
3. Final temperature = 100 + 96 = 196 °C.

Teacher's Note:
1. A temperature difference is the same in kelvin and in degrees Celsius.
2. Helium is monatomic, so \( C_v=\tfrac32R \).

Answer: (d) 1500 J

Explanation:
1. The volume decreases by \( \Delta V=1.5\times10^{-2} \) m³.
2. Work done on the gas = \( P\,|\Delta V|=10^5\times1.5\times10^{-2}=1500 \) J.

Teacher's Note:
1. The question asks for work done on the gas, which is positive in a compression.
2. Work done by the gas would be −1500 J.

Answer: (c) 12 K

Explanation:
1. Isochoric: \( Q=n\tfrac32R(20) \).
2. Isobaric: \( Q=n\tfrac52R\,\Delta T' \).
3. \( \Delta T'=20\times\dfrac{3/2}{5/2}=12 \) K.

Teacher's Note:
1. At constant pressure part of the heat goes into work, so the temperature rise is smaller.
2. Compare \( C_v \) and \( C_p \) directly.

Answer: (b) 0.0091 J

Explanation:
1. Initial volume = 1 cm³ and final volume = 1.091 cm³, so \( \Delta V=0.091 \) cm³ \( =0.091\times10^{-6} \) m³.
2. \( W=P\Delta V=10^5\times0.091\times10^{-6}=0.0091 \) J.

Teacher's Note:
1. Convert cm³ to m³ by multiplying by \( 10^{-6} \).
2. Water expands on freezing, so the work done by it is positive.

Answer: (a) 765 J

Explanation:
1. Over a cycle \( \Delta U=0 \), so \( \sum W=\sum Q \).
2. \( \sum Q=5960-5585-2980+3645=1040 \) J.
3. \( 2200-825-1100+W_4=1040 \), giving \( W_4=765 \) J.

Teacher's Note:
1. Keep the signs of heat and work while adding.
2. The net work in a cycle equals the net heat absorbed.

Answer: (d) \( \dfrac{208}{1921} \)

Explanation:
1. Net work = \( \sum Q=1040 \) J.
2. Heat absorbed = \( 5960+3645=9605 \) J (only positive Q).
3. \( \eta=\dfrac{1040}{9605}=\dfrac{208}{1921}\approx10.8\% \).

Teacher's Note:
1. Only the steps in which the gas takes in heat count in the denominator.
2. Efficiency is always less than 1.

Answer: (c) 110 J

Explanation:
1. Work done by the gas = \( P\,A\,x=10^5\times40\times10^{-4}\times0.25=100 \) J.
2. \( Q=\Delta U+W=10+100=110 \) J.

Teacher's Note:
1. Convert 40 cm² to \( 40\times10^{-4} \) m².
2. The first law is Q = ΔU + W with W done by the gas.

Answer: (b) \( 33600+0.02 \) J

Explanation:
1. Heat supplied \( Q=2\times4200\times4=33600 \) J.
2. Volume change: \( 2/1000-2/999.9=-2.0\times10^{-7} \) m³, so \( W=P\Delta V=-0.02 \) J.
3. \( \Delta U=Q-W=33600+0.02 \) J.

Teacher's Note:
1. Water contracts between 0 °C and 4 °C, so the work done by it is slightly negative.
2. The correction is tiny but has a definite sign.

Answer: (a) 7

Explanation:
1. \( \gamma=1+\dfrac2f \).
2. \( \dfrac97-1=\dfrac27=\dfrac2f \), so f = 7.

Teacher's Note:
1. Use \( \gamma=1+2/f \).
2. A gas with 7 degrees of freedom has a vibrational mode active.

Answer: (d) 2.5R

Explanation:
1. T = 400 K and \( U=nC_vT \), so \( C_v=\dfrac{1200R}{2\times400}=1.5R \).
2. \( C_p=C_v+R=2.5R \).

Teacher's Note:
1. Convert 127 °C to 400 K.
2. The gas is monatomic since \( C_v=\tfrac32R \).

Answer: (c) 3R

Explanation:
1. \( dU=nC_vdT \), so \( dQ=2nC_vdT \).
2. \( C=2C_v=2\times\tfrac32R=3R \).

Teacher's Note:
1. Molar heat capacity is dQ per mole per unit temperature change.
2. Here the work equals the internal energy change.

Answer: (b) 4.2 J/cal

Explanation:
1. \( C_p-C_v=6.90-4.93=1.97 \) cal/mol K, which equals R.
2. \( J=\dfrac{8.3}{1.97}\approx4.2 \) J/cal.

Teacher's Note:
1. R in calories equals \( C_p-C_v \).
2. Dividing joules by calories gives the conversion factor.

Answer: (a) 40 J

Explanation:
1. Work done \( W=P\Delta V=3\times10^5\times200\times10^{-6}=60 \) J.
2. \( \Delta U=Q-W=100-60=40 \) J.

Teacher's Note:
1. Convert cm³ to m³ first.
2. In an isobaric expansion part of the heat is used as work.

Answer: (d) 125/9 J/mol K

Explanation:
1. \( n=\dfrac{P_iV_i}{RT}=\dfrac{60}{(25/3)(400)}=0.018 \) mol.
2. \( P\Delta V=nR\Delta T=60 \) J gives \( \Delta T=400 \) K.
3. \( C_p=\dfrac{Q}{n\Delta T}=\dfrac{100}{0.018\times400}=\dfrac{125}{9} \) J/mol K.

Teacher's Note:
1. Check: \( C_p-C_v=125/9-50/9=25/3=R \).
2. \( C_v=50/9 \) J/mol K for the same gas.

Answer: (c) 0.8 J/mol K

Explanation:
1. \( \Delta U=nC_v\Delta T \).
2. \( C_v=\dfrac{80}{5\times20}=0.8 \) J/mol K.

Teacher's Note:
1. The temperature change is 20 K.
2. At constant volume all the heat goes into internal energy.

Answer: (b) 3/2

Explanation:
1. \( C_v \) of the mixture \( =\dfrac{\tfrac32R+\tfrac52R}{2}=2R \).
2. \( C_p=\dfrac{\tfrac52R+\tfrac72R}{2}=3R \).
3. \( \gamma=\dfrac{3R}{2R}=\dfrac32 \).

Teacher's Note:
1. Average the molar heat capacities, not the γ values.
2. The result lies between 7/5 and 5/3.

Answer: (a) 128

Explanation:
1. Adiabatic: \( P\propto d^{\gamma} \).
2. \( \dfrac{P'}{P}=32^{7/5}=2^{7}=128 \).

Teacher's Note:
1. \( 32=2^5 \), so the power 7/5 gives \( 2^7 \).
2. Density is inversely proportional to volume.

Answer: (d) 32P

Explanation:
1. \( PV^{\gamma}=\text{const} \), so \( P'=P\left(\dfrac{640}{80}\right)^{5/3} \).
2. \( 8^{5/3}=2^5=32 \), so \( P'=32P \).

Teacher's Note:
1. The volume ratio is 8.
2. \( 8^{1/3}=2 \), which makes the power easy.

Answer: (c) 4/9 %

Explanation:
1. \( PV^{\gamma}=\text{const} \) gives \( \dfrac{\Delta P}{P}=-\gamma\dfrac{\Delta V}{V} \).
2. \( \dfrac{|\Delta V|}{V}=\dfrac{1}{\gamma}\times\dfrac23\%=\dfrac23\times\dfrac23\%=\dfrac49\% \).

Teacher's Note:
1. Differentiate \( \ln P+\gamma\ln V=\text{const} \) for small changes.
2. The fractional change in volume is smaller than that in pressure when γ > 1.

Answer: (b) \( 200\sqrt2 \) K

Explanation:
1. \( TV^{\gamma-1}=\text{const} \), so \( T_2=400\times2^{-0.5} \).
2. \( T_2=\dfrac{400}{\sqrt2}=200\sqrt2\approx283 \) K.

Teacher's Note:
1. Adiabatic expansion lowers the temperature.
2. \( \gamma-1=0.5 \) here.

Answer: (a) \( 40(2-\sqrt2) \) J

Explanation:
1. Final pressure \( P_2=4\times10^5/2^{1.5}=\sqrt2\times10^5 \) Pa.
2. \( W=\dfrac{P_1V_1-P_2V_2}{\gamma-1}=\dfrac{40-\sqrt2\times10^5\times2\times10^{-4}}{0.5} \).
3. \( W=2(40-20\sqrt2)=40(2-\sqrt2)\approx23.4 \) J.

Teacher's Note:
1. \( P_1V_1=4\times10^5\times10^{-4}=40 \) J.
2. Use \( W=(P_1V_1-P_2V_2)/(\gamma-1) \) for an adiabatic process.

Answer: (d) 800 kPa, 100 K

Explanation:
1. Slow compression through weakly conducting walls keeps the gas at the surrounding temperature, so it is isothermal: T = 100 K.
2. \( P_2=200\times\dfrac{800}{200}=800 \) kPa.

Teacher's Note:
1. Slow means there is time for heat exchange.
2. A sudden compression would be adiabatic instead.

Answer: (c) 1600 kPa, 200 K

Explanation:
1. A sudden compression leaves no time for heat exchange, so it is adiabatic.
2. \( P_2=200\times4^{1.5}=1600 \) kPa.
3. \( T_2=100\times4^{0.5}=200 \) K.

Teacher's Note:
1. Adiabatic compression raises both pressure and temperature more than the isothermal one.
2. The volume ratio is 4.

Answer: (b) 112 J

Explanation:
1. Adiabatic path: \( \Delta U=+322 \) J (work done on the system).
2. Second path: \( Q=50\times4.2=210 \) J absorbed, and \( \Delta U=Q+W_{on} \).
3. \( W_{on}=322-210=112 \) J.

Teacher's Note:
1. Internal energy is a state function, so ΔU is the same for both paths.
2. Work done on the system adds to its internal energy.

Answer: (a) 7R/2

Explanation:
1. For \( PV^n=\text{const} \), \( C=C_v+\dfrac{R}{1-n} \).
2. \( C=\tfrac32R+\dfrac{R}{1/2}=\tfrac32R+2R=\tfrac72R \).

Teacher's Note:
1. Here n = 1/2.
2. The general polytropic formula covers all the standard processes.

Answer: (d) 7.5R

Explanation:
1. First law: \( \Delta U=Q-\tfrac{2Q}{3}=\dfrac Q3 \).
2. \( \Delta U=nC_v\Delta T=\tfrac52nR\Delta T \), so \( n\Delta T=\dfrac{2Q}{15R} \).
3. \( C=\dfrac{Q}{n\Delta T}=\dfrac{15R}{2}=7.5R \).

Teacher's Note:
1. Express ΔT from the internal energy first.
2. The heat capacity is larger than \( C_p \) because most of the heat does work.

Answer: (c) 1000 J

Explanation:
1. \( P=\dfrac{RT}{V}=\dfrac{R}{kT} \) and \( dV=2kT\,dT \).
2. \( W=\int P\,dV=\int2R\,dT=2R\Delta T \).
3. \( W=2\times\dfrac{25}{3}\times60=1000 \) J.

Teacher's Note:
1. Write P in terms of T using the equation of state.
2. The factor k and the temperature cancel in the integral.