The Question is from neet-physics thermodynamics joules-law-external-work, and it was created by the author P.NARASIMHA

Question

A gas expands from volume $V_1$ to $V_2$ against a variable external pressure given by $P = \frac{k}{V}$, where $k$ is a constant. If the initial pressure is $P_1$ and the final pressure is $P_2$, the work done by the gas is:

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Work done by a gas during expansion
	**Related Topics**: Thermodynamics, Integration
	**Prerequisites**: Definition of work, integration
## SOLUTION PATHWAY
**Given Information**:
	* Initial volume: $V_1$
	* Final volume: $V_2$
	* External pressure: $P = \frac{k}{V}$
	* Initial pressure: $P_1$
	* Final pressure: $P_2$

**Method & Steps**:
	1. Define work done by a gas during expansion.
		Reasoning: The work done by a gas during expansion is given by the integral of the external pressure with respect to volume.
		Formula/Rule: $W = - \int_{V_1}^{V_2} P_{ext} dV$
	2. Substitute the given expression for external pressure.
		Working: $W = - \int_{V_1}^{V_2} \frac{k}{V} dV$
		Key Point:  The negative sign indicates work done by the system.
	3. Evaluate the integral.
		Working: $W = -k \int_{V_1}^{V_2} \frac{1}{V} dV = -k [\ln(V)]_{V_1}^{V_2} = -k(\ln(V_2) - \ln(V_1))$
	4. Simplify using logarithm properties.
		Working: $W = -k \ln(\frac{V_2}{V_1})$
		Key Point: Since we are looking for the work done by the gas, we need to take the absolute value, which leads to the positive form.
	5. Express the answer in the correct form.
		Result: $W = k \ln(\frac{V_2}{V_1})$
		Verification: The result matches option C.
 ## OPTION ANALYSIS
[A] :x: Incorrect. This is the work done if the pressure were constant at $P_1$.
[B] :x: Incorrect. This is the work done if the pressure were constant at $P_2$.
[C] ✓: Correct. This is the result obtained by integrating the work expression with the given pressure.
[D] :x: Incorrect. This is the work done if the pressure were the average of $P_1$ and $P_2$.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting the negative sign in front of the integral. [How to avoid]: Remember that work done by the system is negative.
		2.[Error 2]: Incorrect integration of 1/V. [How to avoid]: Remember that the integral of 1/V is ln(V).
	**Quick Tips**:
		1.[Helpful shortcut]: When the pressure is inversely proportional to the volume, the work done always involves a natural logarithm.
		2.[Memory aid]: Remember the formula for work in variable pressure $W = -\int P dV$ and the integration of 1/V.

P.NARASIMHA · Answer

$P_1(V_2 - V_1)$

P.NARASIMHA · Answer

$P_2(V_2 - V_1)$

P.NARASIMHA · Answer

The correct Option is "$k \ln(\frac{V_2}{V_1})$"

P.NARASIMHA · Answer

$\frac{P_1 + P_2}{2} (V_2 - V_1)$

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