Question

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

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

Answer 1

Answer: (C)

Here, $V_{1}=1 \, L=10^{- 3} \, m^{3}, \, V_{2}=3L=3\times 10^{- 3}m^{3}$
$p_{1}=1$ atm $=1.013\times 10^{5}Nm^{- 2}, \, \gamma =1.40, \, W=?$
As changes are adiabatic,
$p_{1}V_{1}^{\gamma }=p_{2}V_{2}^{\gamma }$
$\frac{p_{1}}{p_{2}}=\left(\frac{V_{2}}{V_{1}}\right)^{\gamma }=\left(3\right)^{1.4}=4.6555$
$\therefore \, \, p_{2}=\frac{p_{1}}{4.6555}=\frac{1.013 \times 10^{5}}{4.6555}$
$=0.217\times 10^{5} \, Nm^{- 2}$
Work done, $W =\frac{p_{1} V_{1} - p_{2} V_{2}}{\gamma - 1}$
$W=\frac{1.013 \times 10^{5} \times 10^{- 3} - 0.217 \times 10^{5} \times 3 \times 10^{- 3}}{1.4 - 1}$
$W = 90.5 \,J$