Select the correct formula for isothermal irreversible process for an ideal gas -

Question

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Question Explanation:

Selecting the correct work and heat formulas for isothermal irreversible expansion.

Concept: First Law of Thermodynamics and Work Done.

Solution:

Isothermal Condition: For an ideal gas, internal energy ( $U$ ) depends only on temperature.

Since $Δ T = 0$ , the change in internal energy $Δ U = 0$ .

First Law Application: From $Δ U = Q + W$ , if $Δ U = 0$ , then $Q = - W$ .

Irreversible Work: For an irreversible process against a constant external pressure ( $P_{e x t}$ ), the work done is $W = - P_{e x t} (V_{2} - V_{1})$ .

Ideal Gas Substitution: Using the ideal gas law $V = \frac{n RT}{P}$ , we substitute $V_{1}$ and $V_{2}$ into the expression. Note that for an irreversible process, the final volume $V_{2}$ is reached when the gas pressure equals the constant external pressure ( $P_{2} = P_{e x t}$ ).

1. Work Formula: $W = - P_{e x t} (V_{2} - V_{1})$

2. Heat Formula ( $Q = - W$ ): $Q = P_{e x t} (V_{2} - V_{1})$

3. Volume Substitution: $V_{2} = \frac{n RT}{P _{2}}$ and $V_{1} = \frac{n RT}{P _{1}}$

4. Final Form: $Q = P_{e x t} (\frac{n RT}{P _{2}} - \frac{n RT}{P _{1}})$

Final Answer: Option (c)

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