If one litre of an ideal gas at a pressure of 20 atm expands isothermally and reversibly to a final volume of 'X' L by absorbing 92.12 L.atm heat, 'X' (in L) is
Show Hint
For isothermal reversible expansion of an ideal gas, the work done by the gas is \( w = -nRT \ln\left(\frac{V_2}{V_1}\right) \). Since \( \Delta U = 0 \) for an isothermal process, the heat absorbed by the gas is equal to the work done by the gas, \( q = -w \).
Step 1: Identify the type of process and the relevant formula.
The process is an isothermal reversible expansion of an ideal gas. The heat absorbed (\( q \)) by the system is equal to the work done by the system (\( -w \)):
$$q = -w = nRT \ln\left(\frac{V_2}{V_1}\right)$$
Since \( P_1V_1 = nRT \) for the initial state, we can also write:
$$q = P_1V_1 \ln\left(\frac{V_2}{V_1}\right)$$
Step 2: Substitute the given values.
Given: \( V_1 = 1 \) L, \( P_1 = 20 \) atm, \( q = 92.12 \) L.atm, \( V_2 = X \) L.
Substituting these values into the formula:
$$92.12 = (20 \text{ atm})(1 \text{ L}) \ln\left(\frac{X}{1}\right)$$
$$92.12 = 20 \ln(X)$$
Step 3: Solve for the final volume \( X \).
Divide both sides by 20:
$$\ln(X) = \frac{92.12}{20} = 4.606$$
Take the exponential of both sides to solve for \( X \):
$$X = e^{4.606}$$
$$X \approx 100$$
Final Answer:
\[
\boxed{100 \text{ L}}
\]
