1. At N.T.P. one mole of diatomic gas iscompressed adiabatically to ha

1.	1. At N.T.P. one mole of diatomic gas iscompressed adiabatically to half of its volumey =1.41. The work doen on gas will be[204 = 1.32]1) 1280 J2) 1610 J3) 1824 J4) 2025 J
Answer» DEAR STUDENT, ◆ Answer - W = 1866 J ◆ Explanation - In adiabatic process, (V1/V2)^(γ-1) = T2/T1 Here, V2 = V1/2, T1 = 273 K, γ = 1.41, (2V1/V1)^(1.41-1) = T2/293 T2 = 2^0.41 × 273 T2 = 1.329 × 273 T2 = 362.8 K Work DONE by expansion of gas in adiatic process is - W = nfR∆T/2 W = 1 × 5 × 8.314 × (362.8-273) / 2 W = 1866 J Therefore, work done on gas is 1866 J. Thanks dear...

1. At N.T.P. one mole of diatomic gas iscompressed adiabatically to half of its volumey =1.41. The work doen on gas will be[204 = 1.32]1) 1280 J2) 1610 J3) 1824 J4) 2025 J

Answer»

DEAR STUDENT,

◆ Answer -

W = 1866 J

◆ Explanation -

In adiabatic process,

(V1/V2)^(γ-1) = T2/T1

Here, V2 = V1/2, T1 = 273 K, γ = 1.41,

(2V1/V1)^(1.41-1) = T2/293

T2 = 2^0.41 × 273

T2 = 1.329 × 273

T2 = 362.8 K

Work DONE by expansion of gas in adiatic process is -

W = nfR∆T/2

W = 1 × 5 × 8.314 × (362.8-273) / 2

W = 1866 J

Therefore, work done on gas is 1866 J.

Thanks dear...

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1. At N.T.P. one mole of diatomic gas iscompressed adiabatically to half of its volumey =1.41. The work doen on gas will be[204 = 1.32]1) 1280 J2) 1610 J3) 1824 J4) 2025 J​

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1. At N.T.P. one mole of diatomic gas iscompressed adiabatically to half of its volumey =1.41. The work doen on gas will be[204 = 1.32]1) 1280 J2) 1610 J3) 1824 J4) 2025 J