JEE Physics Thermodynamics MCQs Set 22

Answer: (c) 130 J

Explanation:
1. Over the cycle \( \Delta U=0 \), so \( \Delta U_{ab}+\Delta U_{bc}=-\Delta U_{ca}=180 \) J.
2. Heat absorbed along abc \( =250+60=310 \) J, so \( W_{abc}=Q-\Delta U=310-180=130 \) J.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) \( \dfrac75Q \)

Explanation:
1. \( Q=nC_V\Delta T=n\cdot\tfrac52R\Delta T \).
2. At constant pressure \( Q'=n\cdot\tfrac72R\Delta T=\dfrac75Q \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) \( \Delta Q_A\gt \Delta Q_B,\ \Delta U_A=\Delta U_B \)

Explanation:
1. \( \Delta U \) depends only on the end states, so \( \Delta U_A=\Delta U_B \).
2. Work is the area under the curve: \( W_A\gt W_B \). By \( Q=\Delta U+W \), \( \Delta Q_A\gt \Delta Q_B \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) d, a, b, c

Explanation:
1. A vertical line is constant volume (isochoric) and a horizontal line is constant pressure (isobaric).
2. The adiabatic curve \( pV^\gamma= \) const is steeper than the isotherm \( pV= \) const, so b is isothermal and c is adiabatic: d, a, b, c.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) 10 J

Explanation:
1. Net work is the area enclosed: a right triangle with base \( 5-1=4 \) m\(^3\) and height \( 6-1=5 \) Pa.
2. Area \( =\tfrac12\times4\times5=10 \) J; the cycle runs C to A to B to C, so the work done by the gas is positive.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) 291 J

Explanation:
1. Isobaric work \( W=nR\Delta T \).
2. \( W=0.5\times8.31\times70\approx291 \) J.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) \( T_2=(T_1^2T_4)^{1/3};\ T_3=(T_1T_4^2)^{1/3} \)

Explanation:
1. Equal efficiencies mean equal ratios: \( \dfrac{T_2}{T_1}=\dfrac{T_3}{T_2}=\dfrac{T_4}{T_3}=r \), so \( r^3=\dfrac{T_4}{T_1} \).
2. \( T_2=T_1r=T_1^{2/3}T_4^{1/3} \) and \( T_3=T_1r^2=T_1^{1/3}T_4^{2/3} \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) 500 K

Explanation:
1. For a Carnot engine \( W=Q_1\dfrac{T_{hot}-T_{cold}}{T_{hot}} \), and the heat rejected by A is the heat received by B. Heat is proportional to temperature through the chain \( \dfrac{Q_1}{T_1}=\dfrac{Q_2}{T_2}=\dfrac{Q_3}{T_3} \).
2. Equal work: \( Q_1-Q_2=Q_2-Q_3\Rightarrow T_1-T_2=T_2-T_3\Rightarrow T_2=\dfrac{600+400}2=500 \) K.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) 189 K and \(-2.7\) kJ

Explanation:
1. Adiabatic: \( T_2=T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma-1}=300\left(\dfrac12\right)^{2/3}\approx189 \) K.
2. \( \Delta U=nC_V\Delta T=2\times\tfrac32\times8.3\times(189-300)\approx-2.7 \) kJ (the internal energy decreases).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) \( \dfrac{9p_0V_0}{4nR} \)

Explanation:
1. The line is \( p=3p_0-\dfrac{p_0}{V_0}V \), so \( T=\dfrac{pV}{nR}=\dfrac{1}{nR}\left(3p_0V-\dfrac{p_0V^2}{V_0}\right) \).
2. \( \dfrac{dT}{dV}=0\Rightarrow V=1.5V_0,\ p=1.5p_0 \), giving \( T_{max}=\dfrac{2.25p_0V_0}{nR}=\dfrac{9p_0V_0}{4nR} \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) 588 J

Explanation:
1. \( p^3V^5= \) const means \( pV^{5/3}= \) const, a monoatomic gas: \( \Delta U=\tfrac32(p_fV_f-p_iV_i)=\tfrac32(25-100)=-112.5 \) J.
2. Isobaric step work \( =p_i(V_f-V_i)=10^5\times7\times10^{-3}=700 \) J, none in the isochoric step. \( Q=\Delta U+W=-112.5+700\approx588 \) J.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) \( \dfrac{13}{2}p_0V_0 \)

Explanation:
1. Heat is absorbed in AB (isochoric, temperature rises) and BC (isobaric expansion). AB: \( Q=\tfrac32(2p_0-p_0)V_0=\tfrac32p_0V_0 \).
2. BC: \( Q=\tfrac52(2p_0)(2V_0-V_0)=5p_0V_0 \). Total from the source \( =\dfrac{13}{2}p_0V_0 \) (CD and DA reject heat).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) P-4, Q-3, R-2, S-1

Explanation:
1. G: isotherm through F gives \( 32p_0V_0=p_0V_G\Rightarrow V_G=32V_0 \). H: adiabat \( p V^{5/3}= \) const gives \( V_H=32^{3/5}V_0=8V_0 \).
2. P (isobaric \( 32V_0\to V_0 \)): \( 31p_0V_0 \). Q (isobaric \( 32V_0\to8V_0 \)): \( 24p_0V_0 \). R (adiabatic): \( \dfrac{32p_0V_0-8p_0V_0}{2/3}=36p_0V_0 \). S (isothermal): \( 32p_0V_0\ln32=160p_0V_0\ln2 \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (a) \( p_3\gt p_1,\ W\lt 0 \)

Explanation:
1. After the isothermal expansion the pressure is \( p_2\lt p_1 \). The adiabatic compression back to \( V_1 \) raises the temperature, so the gas ends hotter than it started and \( p_3\gt p_1 \).
2. On the p-V diagram the adiabat is steeper than the isotherm, so the compression curve lies above the expansion curve; the work done on the gas exceeds the work done by it and \( W\lt 0 \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) \(-5\) J

Explanation:
1. Work in AB \( =p\Delta V=10\times1=10 \) J and work in BC (constant volume) \( =0 \).
2. Over the cycle \( W_{net}=Q_{net}=5 \) J, so \( W_{CA}=5-10-0=-5 \) J.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) \( \text{O}_2\text{ and He} \)

Explanation:
1. For an adiabatic process \( \dfrac{dp}{dV}=-\gamma\dfrac pV \): the larger the \( \gamma \), the steeper the curve.
2. Plot 2 has the larger \( \gamma \). Helium is monoatomic (\( \gamma=\tfrac53 \)) and oxygen is diatomic (\( \gamma=\tfrac75 \)), so plot 1 is O\(_2\) and plot 2 is He (O\(_2\) and N\(_2\) have the same \( \gamma \); He and Ar too).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (a) \( \left(\dfrac{L_2}{L_1}\right)^{2/3} \)

Explanation:
1. Adiabatic process for a monoatomic gas: \( TV^{\gamma-1}= \) const with \( \gamma=\tfrac53 \), so \( TV^{2/3}= \) const.
2. With V proportional to L: \( T_1L_1^{2/3}=T_2L_2^{2/3}\Rightarrow\dfrac{T_1}{T_2}=\left(\dfrac{L_2}{L_1}\right)^{2/3} \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) 42 K

Explanation:
1. A: constant pressure, \( Q=nC_p\Delta T_A \). B: constant volume, \( Q=nC_V\Delta T_B \).
2. \( \Delta T_B=\dfrac{C_p}{C_V}\Delta T_A=1.4\times30=42 \) K.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) \( \dfrac57 \)

Explanation:
1. Heat at constant pressure \( Q=nC_p\Delta T=\tfrac72nR\Delta T \); the part that raises the internal energy is \( \Delta U=nC_V\Delta T=\tfrac52nR\Delta T \).
2. Fraction \( =\dfrac{C_V}{C_p}=\dfrac57 \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (a) (a), (b) and (c) are all correct

Explanation:
1. Case 1: \( p_2=p_1\dfrac{T_2V_1}{T_1V_2}=1.5p_1 \). Spring force \( =(p_2-p_1)A \) over a displacement with \( Ax=V_1 \): energy \( =\tfrac12(0.5p_1)V_1=\tfrac14p_1V_1 \): (a) true. \( \Delta U=\tfrac32(p_2V_2-p_1V_1)=\tfrac32(3-1)p_1V_1=3p_1V_1 \): (b) true.
2. Case 2: \( p_2=\tfrac43p_1 \), \( Ax=2V_1 \). Work done by the gas \( =p_1(2V_1)+\tfrac12\left(\tfrac13p_1\right)(2V_1)=\tfrac73p_1V_1 \): (c) true.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) (b) and (c) only

Explanation:
1. Ice has a larger volume than the water formed from it, so the system shrinks and the atmosphere does positive work on it: (b) true, (a) false.
2. Heat is absorbed at constant temperature (latent heat) and the work done on the system is positive, so \( \Delta U=Q+W_{on}\gt 0 \): (c) true.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) (a), (b), (c) and (d)

Explanation:
1. For an ideal gas U depends only on T, so \( \Delta U=nC_V\Delta T \) for any process: (a) true; \( \Delta U=0 \) when T is constant: (c) true.
2. Adiabatic means \( Q=0 \): (d) true, and then \( \Delta U=-W \): (b) true.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (a) 2

Explanation:
1. Path iaf: \( 500=U_f-100+200\Rightarrow U_f=400 \) J (no work in the leg ia).
2. \( Q_{ib}=(200-100)+50=150 \) J and \( Q_{bf}=(400-200)+100=300 \) J, so the ratio is 2.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) 4

Explanation:
1. \( T_fV_f^{\gamma-1}=T_iV_i^{\gamma-1} \) with \( \gamma=\tfrac75 \).
2. \( a=\left(\dfrac{V_i}{V_f}\right)^{2/5}=32^{2/5}=4 \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) falls more steeply

Explanation:
1. Isothermal slope \( =-\dfrac pV \); adiabatic slope \( =-\gamma\dfrac pV \).
2. Since \( \gamma\gt 1 \), the adiabat is steeper than the isotherm.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (a) \( 50^\circ\text{C} \)

Explanation:
1. \( Q=mc\Delta T \).
2. \( \Delta T=\dfrac{20000}{1\times400}=50^\circ \)C.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) 0.05 J

Explanation:
1. \( V=\dfrac1{9000}=1.11\times10^{-4} \) m\(^3\) and \( \Delta V=V\gamma\Delta T=1.11\times10^{-4}\times9\times10^{-5}\times50=5\times10^{-7} \) m\(^3\).
2. \( W=p\Delta V=10^5\times5\times10^{-7}=0.05 \) J.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) 600 K

Explanation:
1. A to B is isobaric, so \( V\propto T \).
2. \( T_B=T_A\dfrac{V_B}{V_A}=300\times2=600 \) K.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.