Question

one mole of an ideal gas at an initial temperature of T K does 6 R joules of work adiabatically. If the ratio of specific heats constant volume is 5/3, the final temperature of gas will be ​

Answer 1

Answer:

Δ=μCvΔT and 0=W+ΔU

⇒ΔU = −6R(∴W=6R)

Therefore −6R=1(γ−1R)ΔT=23RΔT

ΔT=−4⇒Tfinal=(T−4)K

Answer 2

Q]____?

=> ( T - 4 ) K

Explanation:

In an adiabatic process,

Q = 0

So, from Ist law of thermodynamics,

∆Q = ∆U + ∆W

As, ∆Q = 0

W = - ∆U = -nCv ∆T

W = -n [ R / γ - 1 ] ( Tf - Ti )

W = nR / γ - 1 ( Tf - Ti ) _[i]

Given, work done,

W = 6R J, n = 1 mol,

R = 8.31 J/mol - K, γ = 5/3, Ti = TK

Substituting given values in Eq [i] we get,

∴ 6R = R / ( 5/3 - 1 ) ( T - Tf )

=> 6R = 3R / 2 ( T - Tf )

=> T - Tf = 4

∴ Tf = ( T - 4 ) K

Thus the final temperature of gas will be ( T - 4 ) K !