JEE Main 2016 · 3 Apr · Shift 1 · Physics Q9

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PhysicsSingle correct+4 / -1Question 9 of 90

'n' moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be :

P-V diagram: a straight line from A (V₀, 2P₀) to B (2V₀, P₀).

\dfrac{9P_0V_0}{4nR}

\dfrac{3P_0V_0}{2nR}

\dfrac{9P_0V_0}{2nR}

\dfrac{9P_0V_0}{nR}

Answer and worked solution

Official answer: A

On the line AB, $P=3P_0-\frac{P_0}{V_0}V$ , so $nRT=PV=3P_0V-\frac{P_0}{V_0}V^2$ . This is greatest at $V=\frac32V_0$ , where $P=\frac32P_0$ :
$T_{max}=\frac{9P_0V_0}{4nR}.$

Topics:Ideal gas equation Thermodynamic processes

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JEE Main 2016 · 3 Apr · Shift 1, Physics question 9