$0.02\, moles$ of an ideal diatomic gas with initial temperature $20^{\circ} C$ is compressed from $1500 \,cm ^{3}$ to $500 \,cm ^{3}$. The thermodynamic process is such that $p V^{2}=\beta$, where $\beta$ is a constant. Then, the value of $\beta$ is close to (the gas constant, $R=8.31 \,J / K / mol$ ).
KVPY 2019, Advanced
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$(a)$ Process equation is
$p V^{2}=\beta$ ................$(i)$
As gas is ideal, it obeys gas equation,
$p V=n R T$ .................$(ii)$
From Eqs. $(i)$ and $(ii)$, gives
$\left(n R^{\prime} T\right) \cdot V=\beta$
Here, $n=0.02\, moles$,
$R=8.31 \,JK ^{-1} mol ^{-1}$,
$T=20^{\circ} C +273=293 \,K$
and $V=1500 \,cm ^{3}=1.5 \times 10^{-3} \,m ^{3}$
$\therefore \quad \beta=0.02 \times 8.31 \times 293 \times 1.5 \times 10^{-3}$
$=7.3 \times 10^{-2} \,Pa - m ^{6}$
$\approx 7.5 \times 10^{-2} \,Pa \cdot m ^{6}$
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