39. Two moles of an ideal monoatomic gas at27°C occupies a volume of

1.	39. Two moles of an ideal monoatomic gas at27°C occupies a volume of V.If the gas isexpanded adiabatically to the volume 2V, thenthe work done by the gas will be (y = 5/3)1) -2767.23J 2) 2767.23J3) 2500J4) -2500
Answer» 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...

39. Two moles of an ideal monoatomic gas at27°C occupies a volume of V.If the gas isexpanded adiabatically to the volume 2V, thenthe work done by the gas will be (y = 5/3)1) -2767.23J 2) 2767.23J3) 2500J4) -2500

Answer»

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...

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39. Two moles of an ideal monoatomic gas at27°C occupies a volume of V.If the gas isexpanded adiabatically to the volume 2V, thenthe work done by the gas will be (y = 5/3)1) -2767.23J 2) 2767.23J3) 2500J4) -2500​

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39. Two moles of an ideal monoatomic gas at27°C occupies a volume of V.If the gas isexpanded adiabatically to the volume 2V, thenthe work done by the gas will be (y = 5/3)1) -2767.23J 2) 2767.23J3) 2500J4) -2500