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)

$\text{P}_{1} = \frac{\text{nRT}}{\text{V}_{1}} = \frac{2 \times \text{0.0821} \times \text{243.6}}{2 0} = \text{2 atm}$
$\Delta \text{S} = \text{n} \text{R} \text{\ell } \text{n} \left(\frac{\left(\text{P}\right)_{1}}{\left(\text{P}\right)_{2}}\right)$
$= 2 \times 2 \text{\ell } \text{n } 2$
$= 2 \times 2 \times \text{0.693}$
$\left[\because \text{\ell } \text{n } 2 \, \text{or } \text{log}_{\text{e}} \, 2 = \text{0.693}\right]$
$= \text{2.77 cal/K}$