Work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from a pressure of 20 atm

Question

Work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from a pressure of 20 atmospheres to a pressure of 10 atmospheres is: (Given: R = 2.0 cal K–1 mol–1)

Accepted Answer

The work done ($w$) during the reversible isothermal expansion of an ideal gas is given by the formula: 
$w = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}ight)$
Since $P_1V_1 = P_2V_2$ for an isothermal process, we can also write this as:
$w = -2.303 nRT \log_{10}\left(\frac{P_1}{P_2}ight)$

Given the values:
Number of moles, $n = 1$ mol
Universal gas constant, $R = 2.0 	ext{ cal K}^{-1} 	ext{mol}^{-1}$
Temperature, $T = 25^\circ	ext{C} = 25 + 273 = 298 	ext{ K}$
Initial pressure, $P_1 = 20 	ext{ atm}$
Final pressure, $P_2 = 10 	ext{ atm}$

Substituting these values into the formula:
$w = -2.303 	imes 1 	imes 2.0 	imes 298 	imes \log_{10}\left(\frac{20}{10}ight)$
$w = -2.303 	imes 596 	imes \log_{10}(2)$
$w = -2.303 	imes 596 	imes 0.3010$
$w \approx -413.14 	ext{ calories}$

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