Question

What is the value of delta g and delta a when 2.5 moles of an ideal gas at 298k expands from 10l to 50l in isothermal reversal process?

Answer 1

delta is denoted by triangle

Answer 2

Hey dear,

● Answer -
∆G = 9969 J
∆A = -9969 J

● Explaination -
# Given-
n = 2.5
T = 298 K
V1 = 10 L
V2 = 50 L

# Solution -
In isothermal reversible process, ∆T = 0 & ∆H = 0.

Work done in isothermal reversible system -
W = -nRTln(V2/V1)
W = -2.5 × 8.314 × 298 × ln(50/10)
W = -9969 J

Value of ∆A -
∆A = W
∆A = -9969 J

Value of ∆G -
∆G = ∆H - W
∆G = 0 - 9969
∆G = 9969 J

Hope this helps you...