JEE Main 2014 · 9 Apr · Shift 1 · Physics Q12

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PhysicsSingle correct+4 / -1Question 12 of 90

The equation of state for a gas is given by $PV=nRT+\alpha V$ , where $n$ is the number of moles and $\alpha$ is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are $T_o$ and $P_o$ respectively. The work done by the gas when its temperature doubles isobarically will be :

\dfrac{P_oT_oR}{P_o-\alpha}

\dfrac{P_oT_oR}{P_o+\alpha}

P_oT_oR\ln2

P_oT_oR

Answer and worked solution

Official answer: A

At constant pressure $P_o$ (one mole), $V=\dfrac{RT}{P_o-\alpha}$ . Doubling $T$ from $T_o$ :
$W=P_o\Delta V=\frac{P_oRT_o}{P_o-\alpha}.$

Topics:Work done by a gas Ideal gas equation

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JEE Main 2014 · 9 Apr · Shift 1, Physics question 12