2 mole of an ideal gas at 27°C temperature is expanded reversibly from 2 lit to 20 lit. Find the entropy change (R = 2 cal/mol K) :-

Question

2 mole of an ideal gas at 27°C temperature is expanded reversibly from 2 lit to 20 lit. Find the entropy change (R = 2 cal/mol K) :- (A) 92.1 (B) 0 (C) 4 (D) 9.2

Tardigrade Admin · Accepted Answer

for isothermal reversible expansion w = q = nRT × 2.303 log (v2/v1) = 2RT × 2.303 log (20/2) = 2 × 2 × T × 2.303 × 1 = 9.2 T Entropy change, Δ S = (q/T) = (9.2T/T) = 9.2 cal .

Q. 2 mole of an ideal gas at $2 7^{\circ} C$ temperature is expanded reversibly from 2 lit to 20 lit. Find the entropy change (R = 2 cal/mol K) :-

92.1

0

4

9.2

for isothermal reversible expansion
$w = q = n RT \times 2.303 lo g \frac{v _{2}}{v _{1}}$
$= 2 RT \times 2.303 lo g \frac{20}{2}$
$= 2 \times 2 \times T \times 2.303 \times 1 = 9.2 T$
Entropy change, $Δ S = \frac{q}{T} = \frac{9.2 T}{T} = 9.2$ cal .

Q. 2 mole of an ideal gas at 27∘C temperature is expanded reversibly from 2 lit to 20 lit. Find the entropy change (R = 2 cal/mol K) :-

92.1

0

4

9.2

for isothermal reversible expansion w=q=nRT×2.303logv1​v2​​ =2RT×2.303log220​ =2×2×T×2.303×1=9.2T Entropy change, ΔS=Tq​=T9.2T​=9.2 cal .

Q. 2 mole of an ideal gas at $2 7^{\circ} C$ temperature is expanded reversibly from 2 lit to 20 lit. Find the entropy change (R = 2 cal/mol K) :-

for isothermal reversible expansion
$w = q = n RT \times 2.303 lo g \frac{v _{2}}{v _{1}}$
$= 2 RT \times 2.303 lo g \frac{20}{2}$
$= 2 \times 2 \times T \times 2.303 \times 1 = 9.2 T$
Entropy change, $Δ S = \frac{q}{T} = \frac{9.2 T}{T} = 9.2$ cal .