Grade 11Physical Chemistry

1mole of NH3 gas(gamma=1.33) at 27°C is allowed to expand adiabatically reversibly so that the final volume becomes 8 times. What will be the work done?

Meghant Pandey

8 Years agoGrade 11

2 Answers

Chhavi Jain

8 Years ago

Dear student,

Given : Rev. adiabatic expansion(q=0) , V_f =8 V_i , gamma= 1.33 , n = 1 , T= 300K

using TV^γ-1 = constant

T₁/ T₂= (V2 / V₁ ) ^ (γ-1)

300/ T₂ = 8 ^ ( 1.33-1)

T₂= 150 K ( approx )

Now, using γ = C_p / C_v

1.33 = 4R / 3R = C_p /C_v

So, by compairing both sides we can say that : C_v = 3R

Using : ΔE= q +w

As, q=0 ; ΔE= w

w = nC_vΔT

w= 1 (3R) ( T₂-T₁ )

w= 3R ( 150-300 )

w= -150R

Chemistry Consultant

Askiitians

Rishi Sharma

5 Years ago

Dear Student,
Please find below the solution to your problem.

Given : Rev. adiabatic expansion(q=0) , Vf=8 Vi , gamma= 1.33 , n = 1 , T= 300K
using TVγ-1= constant
T1/ T2= (V2 / V1)^(γ-1)
300/ T2= 8^ ( 1.33-1)
T2= 150 K ( approx )
Now, usingγ = Cp/ Cv
1.33 = 4R / 3R = Cp/Cv
So, by compairing both sides we can say that : Cv= 3R
Using : ΔE= q +w
As, q=0 ;ΔE= w
w = nCvΔT
w= 1 (3R) ( T2-T1)
w= 3R ( 150-300 )
w= -150R

Thanks and Regards