Question

Consider the figure provided. 1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at $18^\circ \text{C}$. If the piston is moved to position B, keeping the temperature unchanged, then 'x' $\text{L atm}$ work is done in this reversible process. $x =$ ______ $\text{L atm}$. (nearest integer)
$[ \text{Given: Absolute temperature} = ^\circ \text{C} + 273.15, \, R = 0.08206 \, \text{L atm mol}^{-1} \text{K}^{-1} ]$

Answer 1

Work done ($W$) in an isothermal reversible expansion of an ideal gas is given by:
$W = -nRT \ln \left( \frac{V_2}{V_1} \right)$
Given:
$n = 1 \, \text{mol}, \quad T = 18^\circ \text{C} = 18 + 273.15 = 291.15 \, \text{K}$
$V_1 = 10 \, \text{L}, \quad V_2 = 100 \, \text{L}$
Substitute the values:
$W = -1 \times 0.08206 \times 291.15 \times \ln \left( \frac{100}{10} \right)$
$W \approx -1 \times 0.08206 \times 291.15 \times \ln(10)$
$W \approx -55.0128 \, \text{L atm}$
The work done by the system is approximately $-55 \, \text{L atm}$ (rounded to nearest integer).

Answer 2

Work done ($W$) in an isothermal reversible expansion of an ideal gas is given by:
$W = -nRT \ln \left( \frac{V_2}{V_1} \right)$
Given:
$n = 1 \, \text{mol}, \quad T = 18^\circ \text{C} = 18 + 273.15 = 291.15 \, \text{K}$
$V_1 = 10 \, \text{L}, \quad V_2 = 100 \, \text{L}$
Substitute the values:
$W = -1 \times 0.08206 \times 291.15 \times \ln \left( \frac{100}{10} \right)$
$W \approx -1 \times 0.08206 \times 291.15 \times \ln(10)$
$W \approx -55.0128 \, \text{L atm}$
The work done by the system is approximately $-55 \, \text{L atm}$ (rounded to nearest integer).