One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (Δ S surr ) in JK -1 is- (1 L atm =101.3 J )

Question

One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (Δ S surr ) in JK -1 is- (1 L atm =101.3 J ) (A) 5.763 (B) 1.013 (C) -1.013 (D) -5.763

Tardigrade Admin · Accepted Answer

W.D. =-P text ex (V2-V1) W.D. =-3(2-1)=-3 × 101.3=-303.9 J Δ E = q + W Δ E =0 for isothermal process q=-w We know Δ S=(q/T) Δ S text system =+(303.9/300)=+1.013 J / K ∴ Δ S text Surrounding =-Δ S text system =-1.013 J / K

Q. One mole of an ideal gas at $300 K$ in thermal contact with surroundings expands isothermally from $1.0 L$ to $2.0 L$ against a constant pressure of $3.0 a t m$ . In this process, the change in entropy of surroundings $(Δ S_{s u rr})$ in $J K^{- 1}$ is- (1 L atm $= 101.3 J)$

5.763

1.013

-1.013

-5.763

W.D. $= - P_{ex} (V_{2} - V_{1})$
W.D. $= - 3 (2 - 1) = - 3 \times 101.3 = - 303.9 J$
$Δ E = q + W {Δ E = 0$ for isothermal process $}$
$q = - w$
We know $Δ S = \frac{q}{T}$
$Δ S_{system} = + \frac{303.9}{300} = + 1.013 J / K$
$∴ Δ S_{Surrounding} = - Δ S_{system}$
$= - 1.013 J / K$

Q. One mole of an ideal gas at 300K in thermal contact with surroundings expands isothermally from 1.0L to 2.0L against a constant pressure of 3.0atm. In this process, the change in entropy of surroundings (ΔSsurr​) in JK−1 is- (1 L atm =101.3J)