IIT JEE past question Thermodynamics 4

Adiabatic = Reversible OR Irreversible ??

TV^5/3-1 = T2 ((2)V)^5/3-1

T = T2(2^2/3)

300/(2^2/3) = T2

2R/(2/3) (T2-T)

Point!

its not given that Constant Ext. Pressure or nethin lyk dat,

Then v can take REVERSIBLE, jus lyk IITIM... took

hmm.......

T1=300K. V1=V. V2=2V. T2=?

mnoatomic => Cv=3/2, Cp=5/2 , γ=5/3

TV^γ-1 = constant.

300.V^2/3 = T2. (2V)^2/3

=> [300/T2] = ³√4

=> T2 = 300/ ³√4 K.

ΔU= nCvΔT. = 2X3/2X (300 - 300/³√4)

= 3.(300 - 300/³√4)

work done = nR(T1-T2)/(γ-1) all the terms we know already.

FOR TEMPERATURE USE
T₁V_I^γ-1=T₂V₂^γ-1

WE GET P₁=2^3/5P₂
AND 2^2/5T₁=T₂

T₁=300K

SO

300 X 2^2/5=T₂

ΔE =nC_vdt

=3/2 x 2 x R x (300 X 2^2/5-300)

as it is adiabatic so ΔQ=0

by applying ΔE=ΔQ-ΔW

we can find the work done

iska answer aise nahin karna kya ????