One mole of an ideal gas at 300 K and 20 atm expands to 2 atm under isothermal and reversible conditions. The work done by the gas is \(-x\ \text{kJ mol}^{-1}\). The value of \(x\) is \((R=8.3\ \text{J K}^{-1}\ \text{mol}^{-1})\)
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Remember the value \(2.303 RT\) at 300 K is approximately \(5700 \text{ J}\) or \(5.7 \text{ kJ}\). Since \(\log(P_1/P_2) = \log(10) = 1\), the answer is directly related to this factor.