Question

$22\times {10}^{-3}$kg of $C{O}_2$ undergo reversible expansion at ${27}^0$C from 2 lit to 20 lit. The work done is nearly?

• A. -1382 cal
• B. +691 cal
• C. -300 cal
• D. -691 cal

Answer 1

Answer: (D)

-691 cal

Solution:- (D) $-691cal$

As we know that,

$\text{No. of moles}=\frac{\text{Wt.}}{\text{Mol. Wt.}}$

Given:-

Weight of $C{O}_2=22\times {10}^{-3}kg=22g$

Mol. weight of $C{O}_2=44g$

Therefore,

$\text{No. of moles}\left(n\right)=\frac{22}{44}=0.5\text{mol}$

$T=27℃=(27+273)K=300K$

$V_i=2L$

$V_f=20L$

As the temperature is constant, the process is isothermal.

Now as we know that, work done for an isothermal reversible process is given as-

$W=-2.303nRt\log \frac{V_f}{V_i}$

$\therefore W=-2.303\times 0.5\times 2\times 300\times \log \frac{20}{2}$

$\Rightarrow W=-690.9cal\approx -691cal$

Hence the work done is nearly $-691cal$.