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)

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

Tardigrade Admin · Accepted Answer

P 1=( nRT / V 1)=(2 × 0.0821 × 243.6/20)=2 atm Δ mathrmS= mathrmnR ℓ mathrmn(( mathrmP1/ mathrmP2)) =2 × 2 ℓ n 2 =2 × 2 × 0.693 [∵ ℓ n 2 or log e 2=0.693] =2.77 cal / K

Q. 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 $2 = 0.693$ )

1.385 cal/K

-1.2 cal/K

1.2 cal/K

2.77 cal/K

$P_{1} = \frac{n RT}{V _{1}} = \frac{2 \times 0.0821 \times 243.6}{20} = 2 a t m$
$Δ S = nR ℓ n (\frac{P _{1}}{P _{2}})$
$= 2 \times 2 ℓ n 2$
$= 2 \times 2 \times 0.693$
[ $∵ ℓ n 2$ or $lo g_{e} 2 = 0.693$ ]
$= 2.77 c a l / K$

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