Question

What is the work done by the 0.2 mole of gas at room temperature to double it's volume during isometric process? Take R=2cak mol-1 degree C-1

Answer 1

Answer is-
In an isobaric process we know that,
W = P∆V 
    = P(V2 - V1) 
In the given equation V1 and V2 are initial and final volumes
here V2 = 2V1, putting that in equation,
      = P(2V1 - V1)
      = PV1 
Using Ideal Gas Law,
PV1 = nRT1 
so, W = nRT1
= 0.2×2×27 cal
= 10.8 cal
W = 10.8 cal 

Answer 2

Let V₁ and V₂ be the initial and final volumes of the gas.
Given that, V₂ = 2V₁
n = 0.2 moles
T = 27°C = 300K
R = 2 cal/mol/°C
So, for isometric process,
Work done = PΔV
Work done = P (V₂ - V₁)
Work done = P (2V₁ - V₁)  , Using (1)
Work done = PV₁
Using Ideal Gas Law,
Work done = nRT
Substitute values, we get:
Work done = (0.2) (2) (300)
Work done = 120 cal
This is the required answer. 
Thanks.