Heat & Thermodynamics — JEE Advanced · 2015 · 24 May 2015 · Paper 2 Question with Solution

An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature

T_{1}

, pressure

P_{1}

and volume

V_{1}

and the spring is in its relaxed state. The gas is then heated very slowly to temperature

T_{2}

, pressure

P_{2}

and volume

V_{2}

. During this process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct statement(s) is(are)

(A)If $V_{2} = 2 V_{1}$ and $T_{2} = 3 T_{1}$ , then the energy stored in the spring is $\frac{1}{4} P_{1} V_{1}$ ✓ Correct
(B)If $V_{2} = 2 V_{1}$ and $T_{2} = 3 T_{1}$ , then the change in internal energy is $3 P_{1} V_{1}$ ✓ Correct
(C)If $V_{2} = 3 V_{1}$ and $T_{2} = 4 T_{1}$ , then the work done by the gas is $\frac{7}{3} P_{1} V_{1}$ ✓ Correct
(D)If $V_{2} = 3 V_{1}$ and $T_{2} = 4 T_{1}$ , then the heat supplied to the gas is $\frac{17}{6} P_{1} V_{1}$

Answer: (A), (B), (C)

Solution

The piston experiences atmospheric pressure

P_{0} = P_{1}

and spring force. At equilibrium:

P = P_{1} + \frac{K x}{A}

.

(A) For

V_{2} = 2 V_{1}

and

T_{2} = 3 T_{1}

: Using ideal gas law,

P_{2} = \frac{3 P _{1}}{2}

. From equilibrium:

\frac{3 P _{1}}{2} = P_{1} + \frac{K x}{A}

, giving

K x = \frac{P _{1} A}{2}

. Since

V_{2} - V_{1} = A x

, we have

x = \frac{V _{1}}{A}

. Energy stored:

\frac{1}{2} K x^{2} = \frac{P _{1} A}{4} \cdot \frac{V _{1}}{A} = \frac{P _{1} V _{1}}{4}

. ✓

(B) Change in internal energy:

Δ U = \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1}) = \frac{3}{2} (\frac{3 P _{1}}{2} \cdot 2 V_{1} - P_{1} V_{1}) = 3 P_{1} V_{1}

. ✓

(C) For

V_{2} = 3 V_{1}

and

T_{2} = 4 T_{1}

P_{2} = \frac{4 P _{1}}{3}

. Work done by gas equals work against atmosphere and spring:

W = P_{1} A \cdot \frac{2 V _{1}}{A} + \frac{1}{2} \cdot \frac{P _{1} A}{3} \cdot \frac{2 V _{1}}{A} = 2 P_{1} V_{1} + \frac{P _{1} V _{1}}{3} = \frac{7 P _{1} V _{1}}{3}

. ✓

(D) Heat supplied:

Q = W + Δ U = \frac{7 P _{1} V _{1}}{3} + \frac{3}{2} (4 P_{1} V_{1} - P_{1} V_{1}) = \frac{7 P _{1} V _{1}}{3} + \frac{9 P _{1} V _{1}}{2} = \frac{41 P _{1} V _{1}}{6}

. ✗

**⚠ Working gives

\frac{41 P _{1} V _{1}}{6}

for option (D); source key states (A,B,C). Flagged for teacher review.**

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Heat & Thermodynamics

Solution

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