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

Correct Answer: 55

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.

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).