JEE Physics Thermodynamics MCQs Set 24

Answer: (c) \( \beta=\dfrac1p\ \text{(a rectangular hyperbola)} \)

Explanation:
1. \( pV= \) const gives \( p\,dV+V\,dp=0\Rightarrow\dfrac{dV}{dp}=-\dfrac Vp \).
2. \( \beta=-\dfrac1V\left(-\dfrac Vp\right)=\dfrac1p \): a decreasing curve, a rectangular hyperbola.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (d) the temperature will decrease

Explanation:
1. First law: \( dU=dQ-dW=dQ\lt 0 \).
2. For an ideal gas \( dU=nC_VdT \), so \( dT\lt 0 \): the temperature decreases (and with \( dW=0 \) the volume is constant).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (b) 513 K

Explanation:
1. Fixed partition: each gas heats or cools at constant volume, and no heat is lost: \( 2\left(\tfrac32R\right)(700-T)=2\left(\tfrac52R\right)(T-400) \).
2. \( 1050-1.5T=2.5T-1000\Rightarrow T=\dfrac{2050}{4}\approx513 \) K.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (c) \( -100R \)

Explanation:
1. With equal and constant pressure on both sides, heat flows at constant pressure: \( 2\left(\tfrac52R\right)(700-T)=2\left(\tfrac72R\right)(T-400)\Rightarrow T=525 \) K.
2. Work by monoatomic gas \( =nR\Delta T=2R(525-700)=-350R \); by diatomic gas \( =2R(525-400)=250R \). Total \( =-100R \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (a) \( \left(\dfrac{p_0\pi R^2}{p_0\pi R^2-Mg}\right)(2L) \)

Explanation:
1. At sealing the trapped air is at \( p_0 \) with length 2L. In equilibrium atmosphere pushes the piston up, while the gas and the weight push it down: \( p_0\pi R^2=p\pi R^2+Mg\Rightarrow p=p_0-\dfrac{Mg}{\pi R^2} \).
2. Isothermal (conducting cylinder): \( p_0(2L)=p\,x\Rightarrow x=\dfrac{p_0\pi R^2}{p_0\pi R^2-Mg}(2L) \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (b) \( \rho g(L_0-H)^2+p_0(L_0-H)-L_0p_0=0 \)

Explanation:
1. The air column has length \( L_0-H \), and its lower surface is at depth \( L_0-H \) below the outer water surface, so its pressure is \( p=p_0+\rho g(L_0-H) \).
2. Isothermal: \( p_0L_0=[p_0+\rho g(L_0-H)](L_0-H)\Rightarrow\rho g(L_0-H)^2+p_0(L_0-H)-L_0p_0=0 \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (c) A: q; B: p, r; C: p, s; D: q, s

Explanation:
1. A: free expansion, \( Q=W=0 \), \( \Delta U=0 \): temperature constant (q). B: \( T\propto pV\propto1/V \) falls, and the molar heat capacity is \( C=\tfrac32R-R=\tfrac R2\gt 0 \), so \( Q\lt 0 \): (p, r).
2. C: \( T\propto V^{-1/3} \) falls; \( C=\tfrac32R+\dfrac{R}{1-4/3}=-\tfrac32R \) is negative, so \( Q=nC\Delta T\gt 0 \): (p, s). D: \( pV \) rises so T rises, \( W\gt 0 \), \( \Delta U\gt 0 \), \( Q\gt 0 \): (q, s).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (a) (b) and (d) only

Explanation:
1. Monoatomic: \( C_V=1.5R,\ C_p=2.5R \). Diatomic: \( C_V=2.5R,\ C_p=3.5R \).
2. \( C_p-C_V=R \) for both (a false); \( C_p+C_V \): 4R vs 6R (b true); \( \gamma \): 1.67 vs 1.4 (c false); \( C_pC_V \): 3.75R\(^2\) vs 8.75R\(^2\) (d true).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

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

Explanation:
1. (a) Area under the line \( =\tfrac12\left(p+\tfrac p2\right)V=0.75pV \), more than the isothermal work \( pV\ln2=0.69pV \): true. (b) Along the line \( p=\dfrac{3p_0}{2}-\dfrac{p_0}{2V_0}V \), so \( T\propto pV \) is a quadratic function of V: a parabola, true.
2. (c) T is quadratic in V, so the p-T curve is not a hyperbola: false. (d) \( pV \) is the same at A and B but larger in between, so T rises to a maximum and then falls: true.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (c) 9

Explanation:
1. Heat flows from P to A: \( Q_A=\dfrac{KA(400-0)}{\lambda x} \), and from P to B: \( Q_B=\dfrac{KA(400-100)}{(10-\lambda)x} \). Equal masses: \( \dfrac{Q_B}{Q_A}=\dfrac{540}{80}=\dfrac{27}4 \).
2. \( \dfrac{300\lambda}{400(10-\lambda)}=\dfrac{27}4\Rightarrow3\lambda=27(10-\lambda)\Rightarrow\lambda=9 \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (a) The specific heat at constant pressure is greater than that at constant volume

Explanation:
1. At constant pressure the gas does work as it expands, in addition to the rise in internal energy.
2. For an ideal gas \( C_p-C_V=R\gt 0 \), so \( C_p\gt C_V \) always.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (b) \( \dfrac{K}{K+4e\sigma LT_s^3} \)

Explanation:
1. Steady state: conduction along the rod equals radiation from the end: \( \dfrac{KA(T_1-T_2)}{L}=e\sigma A(T_2^4-T_s^4)\approx4e\sigma AT_s^3\Delta T \).
2. With \( T_1-T_2=(T_1-T_s)-\Delta T \): \( \Delta T\left(K+4e\sigma LT_s^3\right)=K(T_1-T_s) \), so the constant is \( \dfrac{K}{K+4e\sigma LT_s^3} \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (c) 1.33 m

Explanation:
1. Constant pressure: \( \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2} \), and with fixed area the volume is proportional to the height.
2. \( h_2=1\times\dfrac{400}{300}=1.33 \) m.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (a) \( 163^\circ\text{C} \)

Explanation:
1. In steady state the heat conducted through the disc equals the radiation loss: \( K\dfrac{T_{oil}-400}{d}=595 \).
2. \( T_{oil}-400=\dfrac{595\times0.01}{0.167}\approx35.6 \) K, so \( T_{oil}\approx435.6 \) K \( \approx163^\circ \)C.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (b) \( \dfrac{Mv_0^2}{3R} \)

Explanation:
1. Per mole the kinetic energy of the bulk motion is \( \tfrac12Mv_0^2 \); on stopping it becomes internal energy (no heat exchange, no work).
2. \( \tfrac12Mv_0^2=\tfrac32R\Delta T\Rightarrow\Delta T=\dfrac{Mv_0^2}{3R} \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (d) \( 300+12.5\,e^{-2KAt_1/(CL)}\ \text{K} \)

Explanation:
1. Newton cooling: \( T-300=100e^{-kt/C} \). At \( t_1 \): \( 50=100e^{-kt_1/C}\Rightarrow\dfrac kC=\dfrac{\ln2}{t_1} \).
2. After \( t_1 \) the rod adds a second loss: rate constant \( \dfrac{k}C+\dfrac{KA}{CL} \). For the next \( 2t_1 \): \( T-300=50\exp\left[-2\ln2-\dfrac{2KAt_1}{CL}\right]=12.5\,e^{-2KAt_1/(CL)} \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (a) 2

Explanation:
1. For the mixture \( \gamma=\dfrac{19}{13}\Rightarrow C_V=\dfrac{R}{\gamma-1}=\dfrac{13R}{6} \). \( C_V \) of A \( =1.5R \), of B \( =2.5R \).
2. With x moles of B: \( \dfrac{1(1.5)+x(2.5)}{1+x}=\dfrac{13}{6}\Rightarrow9+15x=13+13x\Rightarrow x=2 \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (b) 401 m/s

Explanation:
1. Average molar mass \( =\dfrac{1(4)+2(32)}{3}=22.67 \) g/mol \( =0.02267 \) kg/mol.
2. \( v=\sqrt{\dfrac{\gamma RT}{M}}=\sqrt{\dfrac{(19/13)(8.314)(300)}{0.02267}}\approx401 \) m/s.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (d) 4.1 g and 23.9 g

Explanation:
1. \( n=\dfrac{pV}{RT}=\dfrac{10^5\times0.02}{8.314\times300}=0.802 \) mol. Let \( n_1+n_2=0.802 \) and \( 20n_1+40n_2=28 \).
2. \( 20(0.802)+20n_2=28\Rightarrow n_2=0.598 \), \( n_1=0.204 \). Masses: neon \( =4.1 \) g, argon \( =23.9 \) g.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (a) 600 K

Explanation:
1. Initially the spring is relaxed, so \( p_1=p_{atm}=2\times10^5 \) Pa. \( V_2=2.4\times10^{-3}+8\times10^{-3}(0.1)=3.2\times10^{-3} \) m\(^3\).
2. \( p_2=p_{atm}+\dfrac{kx}A=2\times10^5+\dfrac{8000\times0.1}{8\times10^{-3}}=3\times10^5 \) Pa. \( T_2=300\times\dfrac{3\times10^5\times3.2\times10^{-3}}{2\times10^5\times2.4\times10^{-3}}=600 \) K.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (b) 920 J

Explanation:
1. \( \Delta U=\tfrac32(p_2V_2-p_1V_1)=1.5(960-480)=720 \) J.
2. Work done by the gas \( =p_{atm}\Delta V+\tfrac12kx^2=2\times10^5(8\times10^{-4})+\tfrac12(8000)(0.01)=160+40=200 \) J. \( Q=720+200=920 \) J.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (d) 12.5 kJ

Explanation:
1. Isobaric step: \( p=\dfrac{nRT}{V}=\dfrac{2(8.314)(300)}{0.02}=2.49\times10^5 \) Pa, \( W_1=p\Delta V=2.49\times10^5\times0.02=4.99 \) kJ.
2. Adiabatic step: \( W_2=nC_V(T_2-T_3)=2(1.5)(8.314)(300)=7.48 \) kJ. Total \( =12.5 \) kJ.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (a) 75.4 cm of Hg

Explanation:
1. Total tube length \( =46+44.5+5=95.5 \) cm, so each end originally had \( 45.25 \) cm of air. Boyle's law: \( p_{up}=\dfrac{45.25p}{46} \), \( p_{low}=\dfrac{45.25p}{44.5} \).
2. The mercury column has vertical height \( 5\cos60^\circ=2.5 \) cm, so \( p_{low}-p_{up}=2.5 \): \( 45.25p\left(\dfrac1{44.5}-\dfrac1{46}\right)=2.5\Rightarrow p\approx75.4 \) cm of Hg.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (b) 83.7 cm of Hg

Explanation:
1. Total moles are conserved: \( \dfrac{2p_0V}{273}=pV\left(\dfrac1{273}+\dfrac1{335}\right) \).
2. \( p=\dfrac{2p_0(335)}{273+335}=\dfrac{2\times76\times335}{608}\approx83.7 \) cm of Hg.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (d) \( \dfrac{207}{16}T_0 \)

Explanation:
1. The total volume is fixed at \( 2V_0 \): \( V_L=2V_0-\dfrac{8}{27}V_0=\dfrac{46}{27}V_0 \).
2. \( T_L=T_0\dfrac{p_fV_L}{p_0V_0}=T_0\times\dfrac{243}{32}\times\dfrac{46}{27}=\dfrac{207}{16}T_0\approx12.9T_0 \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (a) \( -\dfrac{15}{8}RT_0 \)

Explanation:
1. Adiabatic: \( W=-\Delta U=-nC_V\Delta T \).
2. \( W=-\tfrac32R\left(\dfrac94T_0-T_0\right)=-\dfrac{15}{8}RT_0 \): the right gas has \( \dfrac{15}8RT_0 \) of work done on it.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (b) 36 m/s

Explanation:
1. The bulk kinetic energy of one mole becomes internal energy of a diatomic gas: \( \tfrac12Mv_0^2=nC_V\Delta T=\tfrac52R\Delta T \).
2. \( v_0^2=\dfrac{5R\Delta T}{M}=\dfrac{5(8.314)(1)}{0.032}=1299\Rightarrow v_0\approx36 \) m/s.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.