Question

The change in entropy of $2$ moles of an ideal gas upon isothermal expansion at $243.6\, K$ from $20\, L$ to the state where pressure becomes $1$ atm is
(Given: ln $2 = 0.693$)

• A. 1.385 cal/K
• B. -1.2 cal/K
• C. 1.2 cal/K
• D. 2.77 cal/K

Answer 1

Answer: (D)

$P _{1}=\frac{ nRT }{ V _{1}}=\frac{2 \times 0.0821 \times 243.6}{20}=2 atm$
$\Delta \mathrm{S}=\mathrm{nR} \ell \mathrm{n}\left(\frac{\mathrm{P}_1}{\mathrm{P}_2}\right)$
$=2 \times 2 \ell n 2$
$=2 \times 2 \times 0.693$
[$\because \ell n 2$ or $\log _{ e } 2=0.693$]
$=2.77\, cal / K$