One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1 L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is: [R=0.082 L atm mol -1 K -1=8.3 J mol -1 K -1]

Question

One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1 L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is: [R=0.082 L atm mol -1 K -1=8.3 J mol -1 K -1] (A) 0 (B) R ℓ n(24.6) (C) R ℓ n (2490) (D) (3/2) R ℓ(24.6)

Tardigrade Admin · Accepted Answer

Δ S = nR ℓ n (( V f / V i )) = R ℓ n (( P i / P f )) = R ℓ n ((300 R /1 L × 1 atm )) = R ℓ n (24.6)

$0$

$R ℓ n (24.6)$

R $ℓ n (2490)$

$\frac{3}{2} R ℓ (24.6)$

$Δ S = n R ℓ n (\frac{V _{f}}{V _{i}})$
$= R ℓ n (\frac{P _{i}}{P _{f}})$
$= R ℓ n (\frac{300 R}{1 L \times 1 a t m})$
$= R ℓ n (24.6)$

0

Rℓn(24.6)

R ℓn(2490)

23​Rℓ(24.6)

ΔS=nRℓn(Vi​Vf​​) =Rℓn(Pf​Pi​​) =Rℓn(1L×1atm300R​) =Rℓn(24.6)

$0$

$R ℓ n (24.6)$

R $ℓ n (2490)$

$\frac{3}{2} R ℓ (24.6)$

$Δ S = n R ℓ n (\frac{V _{f}}{V _{i}})$
$= R ℓ n (\frac{P _{i}}{P _{f}})$
$= R ℓ n (\frac{300 R}{1 L \times 1 a t m})$
$= R ℓ n (24.6)$