Question

A monoatomic ideal gas ($C_V=\frac{3}{2}R$) is allowed to expand adiabatically and reversibly from initial volume of 8L at 300 K to a volume of $V_2$ at 250 K. $V_2$ is:

(Given ${\left(4.8\right)}^{1/2}$ = 2.2)

• A. 10.5 L
• B. 23 L
• C. 8.5 L
• D. 50.5 L

Answer 1

Answer: (A)

10.5 L

$T{V}^{\gamma -1}=constant$
$\gamma =\frac{5}{3}$

Therefore,$\gamma -1=\frac{2}{3}$
$300\times {\left(8\right)}^{2/3}=250\times {\left(V_2\right)}^{2/3}⟹{\left(V_2\right)}^{2/3}=4.8$
$⟹V_2={\left(4.8\right)}^{3/2}\approx 4.8\times 2.2=10.5L$