In adiabatic irreversible expansion of an ideal gas against a constant external pressure P 2- the P - V work is given by -A- - W irr - nR T 2- T 1 P 2- P 1B- - W irr - nR T 2- T 1 p 2- P 1C- - W irr - nRT 11- P 2- p 1D- -Wirr -n R T21-P2-P1

The correct option is A W_irr = nR (T_2 T_1 P_2/P_1) C_v for diatomic ideal gas = 3/2 RT + R = 5/2 R mol^ 1 C_P for diatomic gas = C_v + R = 5/2 R + R = 7/ ...

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Byju's Answer

Standard XII

Chemistry

Heat Capacity at Constant Volume

In adiabatic ...

Question

In adiabatic irreversible expansion of an ideal gas against a constant external pressure $P_{2},$ the P - V work is given by ?

$- W_{i r r} = n R (T_{2} - \frac{T_{1} P_{2}}{P_{1}})$

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$- W_{i r r} = n R (T_{2} + \frac{T_{1} P_{2}}{P_{1}})$

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$- W_{i r r} = n R T_{1} (1 - \frac{P_{2}}{P_{1}})$

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$- W_{i r r} = n R T_{2} (1 - \frac{P_{2}}{P_{1}})$

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Solution

The correct option is A
$- W_{i r r} = n R (T_{2} - \frac{T_{1} P_{2}}{P_{1}})$

$C_{v}$ for diatomic ideal gas $= \frac{3}{2} R T + R = \frac{5}{2} R m o l^{- 1}$
$C_{P}$ for diatomic gas $= C_{v} + R = \frac{5}{2} R + R = \frac{7}{2} R m o l^{- 1}$
Enthalpy increase of 2 moles $= \frac{7}{2} R \times 2 = 7 R$

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