Question

$0.02$ 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, $R=8.31\, J / K / mol )$.

• A. $7.5 \times 10^{-2}Pa-m^{6}$
• B. $1.5 \times10^{2}Pa - m^{6}$
• C. $3 \times 10^{-2}Pa - m^{6}$
• D. $2.0 \times 10^{1} Pa - m^{6}$

Answer 1

Answer: (A)

Process equation i s
$pV^{2}=\beta\,\,\,...(1)$
As gas is ideal, it obeys gas equation,
$p V=n R T\,\,\,\,...(ii)$
From Eqs. $(i)$ and $(ii), $ gives
$\left(nRT\right) \cdot V=\beta$
Here, $n = 0.02$ moles,
$R=8.31\,JK^{-1}\,mol^{-1}$
$ T=20^{\circ}C+273=293\,K$
and $V=1500 \,cm^{3}=1.5\times10^{-3}\, m^{3}$
$\therefore \beta=0.02\times8.31\times293\times1.5\times10^{-3}$
$=7.3\times10^{-2} \, Pa-m^6$
$\approx7.5\times10^{-2} Pa-m^{6}$