The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is

(Given R = 2.0 cal K^–1 mol^–1)

The correct option is –413.14 calories.

To find the work done during the reversible isothermal expansion of an ideal gas, we use the thermodynamic formula for work done (W) during a reversible isothermal process:
$W=-2.303nRT{\log }_{10}\left(\frac{P_1}{P_2}\right)$

Let's identify the given values from the problem:
• Number of moles ($n$) = 1 mole of hydrogen gas
• Temperature ($T$) = 25°C = 25 + 273.15 = 298.15 K (approximated to 298 K)
• Initial pressure ($P_1$) = 20 atm
• Final pressure ($P_2$) = 10 atm
• Gas constant ($R$) = 2.0 cal K^-1 mol^-1

Now, substitute these values into the formula:
$W=-2.303\times 1\times 2.0\times 298\times {\log }_{10}\left(\frac{20}{10}\right)$

Simplify the expression inside the logarithm:
${\log }_{10}\left(\frac{20}{10}\right)={\log }_{10}\left(2\right)\approx 0.3010$

Now calculate the work done:
$W=-2.303\times 2.0\times 298\times 0.3010$
$W=-4.606\times 298\times 0.3010$
$W=-1372.588\times 0.3010$
$W\approx -413.14\text{calories}$

The negative sign indicates that work is done by the system (gas) during the expansion.