Question

A litre of dry air at STP allowed to expand to a volume of $3L$ under adiabatic conditions. If $\gamma =1.40$ , the work done is $(3^{1.4}=4.6555)$

• A. 48 J
• B. 60.7 J
• C. 90.5 J
• D. 100.8 J

Answer 1

Answer: (C)

Here, $V_1=1L=10^{-3}{m}^3,V_2=3L=3\times 10^{-3}{m}^3$
$p_1=1$ atm $=1.013\times 10^{5}N{m}^{-2},\gamma =1.40,W=?$
As changes are adiabatic,
$p_1{V}_1^{\gamma }=p_2{V}_2^{\gamma }$
$\frac{p_1}{p_2}={\left(\frac{V_2}{V_1}\right)}^{\gamma }={\left(3\right)}^{1.4}=4.6555$
$\therefore p_2=\frac{p_1}{4.6555}=\frac{1.013\times 10^5}{4.6555}$
$=0.217\times 10^{5}N{m}^{-2}$
Work done, $W=\frac{p_1{V}_1-p_2{V}_2}{\gamma -1}$
$W=\frac{1.013\times 10^5\times 10^{-3}-0.217\times 10^5\times 3\times 10^{-3}}{1.4-1}$
$W=90.5J$