Question

If 100 L of an ideal gas at a pressure of 2 atm is compressed isothermally and reversibly to a final volume of ‘X’ L releases -460.6 L atm heat, the final volume ‘X’ (in L) is

• A. 1
• B. 10
• C. 100
• D. 2

Hint:

In isothermal compression, the work done is equal to the heat released, and the relationship between initial and final volumes can be derived using the first law of thermodynamics.

Answer 1

The Correct Option is C

For an isothermal and reversible process, the work done \(W\) is given by: \[ W = -nRT \ln\left( \frac{V_f}{V_i} \right) \] where \( n \) is the number of moles, \( R \) is the universal gas constant, \( T \) is the temperature, \( V_f \) is the final volume, and \( V_i \) is the initial volume. We can use the first law of thermodynamics: \[ Q = \Delta U + W \] where \( Q \) is the heat released, and \( \Delta U \) is the change in internal energy. Since the process is isothermal, \( \Delta U = 0 \), so: \[ Q = W \] Given: \[ P_i = 2 \, \text{atm}, \, V_i = 100 \, \text{L}, \, Q = -460.6 \, \text{L atm} \] We use the formula for the work done: \[ W = P_i (V_i - V_f) \] Substitute the given values: \[ -460.6 = 2 \times (100 - X) \] Solve for \( X \): \[ 100 - X = \frac{-460.6}{2} \] \[ 100 - X = -230.3 \] \[ X = 100 + 230.3 = 10 \] Thus, the final volume \( X \) is \( 10 \, \text{L} \). Thus, the correct answer is \( \boxed{10} \).