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) \([ \text{Given: Absolute temperature} = ^\circ \text{C} + 273.15, \, R = 0.08206 \, \text{L atm mol}^{-1} \text{K}^{-1} ]\)
The work done by the gas in a reversible isothermal process can be calculated using the formula: \( W = nRT \ln\left(\frac{V_f}{V_i}\right) \) Where:
\( W \) is the work done.
\( n = 1 \, \text{mol} \) (amount of gas).
\( R = 0.08206 \, \text{L atm mol}^{-1} \text{K}^{-1} \) (universal gas constant).
\( T \) is the absolute temperature in Kelvin. Given \( T = 18^\circ\text{C} + 273.15 = 291.15 \, \text{K} \).
\( V_i = 10 \, \text{L} \) (initial volume).
\( V_f = 100 \, \text{L} \) (final volume).
Substitute the values into the formula: \( W = 1 \times 0.08206 \times 291.15 \times \ln\left(\frac{100}{10}\right) \) \( W = 23.89989 \times \ln(10) \) \( \ln(10) = 2.302 \) (approx.) \( W \approx 23.89989 \times 2.302 \approx 55.028 \) To the nearest integer, \( W = 55 \, \text{L atm} \). This value falls within the expected range of 55, verifying its correctness.
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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).
