JEE Advanced 2020 · Paper 1 · Physics Q16

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PhysicsNumerical+4Question 16 of 54

Consider one mole of helium gas enclosed in a container at initial pressure $P_1$ and volume $V_1$ . It expands isothermally to volume $4V_1$ . After this, the gas expands adiabatically and its volume becomes $32V_1$ . The work done by the gas during isothermal and adiabatic expansion processes are $W_{iso}$ and $W_{adia}$ , respectively. If the ratio $\dfrac{W_{iso}}{W_{adia}} = f\ln 2$ , then $f$ is ______.

Your answer

Answer and worked solution

Official answer: 1.77

One mole of helium, $\gamma=\frac53$ , with $P_1V_1=RT_1$ .

Isothermal, $V_1\to4V_1$ .
$W_{iso}=RT_1\ln4=2RT_1\ln2.$

Adiabatic, $4V_1\to32V_1$ .
$T_2=T_1\left(\frac4{32}\right)^{\gamma-1}=T_1\left(\frac18\right)^{2/3}=\frac{T_1}4,\qquad W_{adia}=\frac{R(T_1-T_2)}{\gamma-1}=\frac{\frac34RT_1}{\frac23}=\frac98RT_1.$

Ratio.
$\frac{W_{iso}}{W_{adia}}=\frac{2\ln2}{9/8}=\frac{16}9\ln2\implies f=\frac{16}9=1.78.$
Answer: 1.78

Topics:Work done by a gas Thermodynamic processes

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