MCQ
$0.02\, moles$ 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, $R=8.31 \,J / K / mol$ ).
  • $7.5 \times 10^{-2} \,Pa - m ^{6}$
  • B
    $1.5 \times 10^{2} \,Pa - m ^{6}$
  • C
    $3 \times 10^{-2} \,Pa \cdot m ^{6}$
  • D
    $2.0 \times 10^{1} \,Pa \cdot m ^{6}$

Answer

Correct option: A.
$7.5 \times 10^{-2} \,Pa - m ^{6}$
a
$(a)$ 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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