Question

$0.02$ moles of an ideal diatomic gas with initial temperature $20^{\circ }C$ is compressed from $1500c{m}^3$ to $500c{m}^3$. 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, $R=8.31J/K/mol)$.

• A. $7.5\times 10^{-2}Pa-m^6$
• B. $1.5\times 10^{2}Pa-m^6$
• 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
$p{V}^2=\beta ...(1)$
As gas is ideal, it obeys gas equation,
$pV=nRT...(ii)$
From Eqs. $(i)$ and $(ii),$ gives
$\left(nRT\right)\cdot V=\beta$
Here, $n=0.02$ moles,
$R=8.31J{K}^{-1}mo{l}^{-1}$
$T=20^{\circ }C+273=293K$
and $V=1500c{m}^3=1.5\times 10^{-3}{m}^3$
$\therefore \beta =0.02\times 8.31\times 293\times 1.5\times 10^{-3}$
$=7.3\times 10^{-2}Pa-m^6$
$\approx 7.5\times 10^{-2}Pa-m^6$