JEE Advanced 2015 · Paper 2 · Physics Q15

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PhysicsMultiple correct+4 / -2Question 15 of 60

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 horizontal cylinder: gas (dotted) on the left, a piston, and a relaxed spring joining the piston to the fixed right wall (hatched).

V_2=2V_1

and

T_2=3T_1

, then the energy stored in the spring is

\dfrac14P_1V_1

V_2=2V_1

and

T_2=3T_1

, then the change in internal energy is

3P_1V_1

V_2=3V_1

and

T_2=4T_1

, then the work done by the gas is

\dfrac73P_1V_1

V_2=3V_1

and

T_2=4T_1

, then the heat supplied to the gas is

\dfrac{17}6P_1V_1

Answer and worked solution

Official answer: A, B, C

Spring-loaded piston. The pressure rises linearly with volume, $P=P_1+\dfrac k{A^2}(V-V_1)$ . From $\dfrac{PV}T$ = constant: $P_2=P_1\dfrac{T_2V_1}{T_1V_2}$ .

Case $V_2=2V_1$ , $T_2=3T_1$ . $P_2=\frac32P_1$ .

(A) Spring energy $=\frac12kx^2=\frac12(P_2-P_1)(V_2-V_1)=\frac14P_1V_1$ . True.
(B) $\Delta U=\frac32nR\Delta T=\frac32P_1V_1\cdot2=3P_1V_1$ . True.

Case $V_2=3V_1$ , $T_2=4T_1$ . $P_2=\frac43P_1$ .

(C) $W=\frac12(P_1+P_2)(V_2-V_1)=\frac12\cdot\frac73P_1\cdot2V_1=\frac73P_1V_1$ . True.
(D) $Q=\Delta U+W=\frac92P_1V_1+\frac73P_1V_1=\frac{41}6P_1V_1$ . False.

(A), (B), (C)

Topics:Work done by a gas First law and internal energy Spring potential energy

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JEE Advanced 2015 · Paper 2, Physics question 15