Question

39. Two moles of an ideal monoatomic gas at
27°C occupies a volume of V.If the gas is
expanded adiabatically to the volume 2V, then
the work done by the gas will be (y = 5/3)
1) -2767.23J 2) 2767.23J
3) 2500J
4) -2500​

Answer 1

Dear Student,

◆ Answer - (1)

W = -2768 J

● Explanation -

In adiabatic process,

(V1/V2)^(γ-1) = T2/T1

Here, V1 = V, V2 = 2V, T1 = 27 °C = 300 K, γ = 5/3,

(V/2V)^(5/3-1) = T2/300

T2 = (1/2)^(2/3) × 300

T2 = 0.63 × 300

T2 = 189 K

Work done by expansion of gas in adiatic process is -

W = nfR∆T/2

W = 2 × 3 × 8.314 × (189-300) / 2

W = -2768 J

Hope this helps you...