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2 mole of ideal gas undergoes a state change from 2atm to 20atm at constant temperature of 300K. What is the change in Gibb’s free energy for this state change in kJ? (Use $R = 8 J K^{- 1} m o l^{- 1}$ and $l n 10 = 2.3$ ) rounding to nearest integer

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answer is 11.

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Detailed Solution

$d G = V d P - S d T$

$\int d G = \int \frac{n R T}{P} d P Δ G = n R T l n \frac{P_{2}}{P_{1}} Δ G = 2 \times 8 \times 300 \times 2.3$

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