Question

One mole ideal gas is compressed
27 C. Its temperature becomes 102 C. The work
done in this process will be :- (y = 1.5)
(1) -625 J
(2) 625 J
(3) 1245 J
(4) -1245 J​

Answer 1

Answer:

The work done in this process will be -1245J

Explanation:

Question:- One mole ideal gas is compressed adiabatically at 27°C. Its temperature becomes 102°C. The work done in this process will be ?

Solution:- From first law of thermodynamics

dq = du + dw

so, we know that ,for adiabatic process , dq = 0

∵ du = -dw = (-nR/γ-1 )ΔT

Given :- n = 1 , γ =1.5 , T₁ =27°c , T₂ = 102°c

R = 8.314J/mol•K

• => - (nR/1.5-1)ΔT [∵ γ = 1.5]

• => - [(1×8.314)/0.5] (T₂-T₁)

• => -(16.62)× (102-27 )

• => -(16.62)×(75) =-1246.5J ≈ -1245J

so, work done in this process will be -1245J

option (4) is correct.