28 g of $N_{2}$ gas at 300 K and 20 atm was allowed to expand isothermally against a constant external pressure of 1 atm. $Δ$ U, q and w for the gas is :

Δ U = 0, q = 2370 J, w = - 23.70 \times 10^{2} J

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Δ U = 2000, q = 23715, w = 24.70 \times 10^{2} J

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Δ U = 0, q = 2374 J, w = - 23.70 \times 10^{2} J

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Δ U = 0, q = 2370 J, w = - 24.70 \times 10^{2} J

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Solution

The correct option is A $Δ U = 0, q = 2370 J, w = - 23.70 \times 10^{2} J$
Ideal gas equation PV=nRT is used to calculate the initial and final volume.
Initial volume $V_{1} = \frac{(28 / 28) \times 0.0821 \times 300}{20} = 1.23 L$
Final volume $= V_{2} \frac{(28 / 28) \times 0.0821 \times 300}{1} = 24.63 L$
Change in volume, $Δ V = (V_{2} - V_{1}) = (24. 63 - 1.23) L = 23.40 \times 10^{- 3} m^{3}$
The external pressure is $P = 1 a t m = 1.013 \times 10^{5} N m^{- 2}$
The work done is the product of pressure and change in volume.
$w = - P \times Δ V = - 1.013 \times 10^{5} \times 23.40 \times 10^{- 3} = - 23.70 \times 10^{2} J$
During an isothermal process, the temperature remains constant. Hence, the internal energy of the system remains constant.

Thus, $Δ U = 0$ .

Hence, $q = - W$

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28 g of N2 gas at 300 K and 20 atm was allowed to expand isothermally against a constant external pressure of 1 atm. ΔU, q and w for the gas is :

28 g of $N_{2}$ gas at 300 K and 20 atm was allowed to expand isothermally against a constant external pressure of 1 atm. $Δ$ U, q and w for the gas is :