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Question

A gas undergoes a process where its pressure increases linearly with volume, $P = aV + b$, where $a$ and $b$ are positive constants. If the initial volume is $V_0$ and the final volume is $2V_0$, what is the work done by the gas during this process?

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Work done by a gas during a thermodynamic process.
 **Related Topics**:  Thermodynamics, Isobaric, Isothermal, and Adiabatic processes.
 **Prerequisites**: Understanding of integration and the formula for work done by a gas ($W = \int_{V_i}^{V_f} P dV$).
## SOLUTION PATHWAY
**Given Information**:
  $P = aV + b$ (linear relationship between pressure and volume)
  $V_i = V_0$ (initial volume)
  $V_f = 2V_0$ (final volume)
  $a$ and $b$ are positive constants.
**Method & Steps**:
 1. Determine the formula for work done.
  Reasoning: The work done by a gas is given by the integral of pressure with respect to volume.
  Formula/Rule: $W = \int_{V_i}^{V_f} P dV$
 2. Substitute the given pressure-volume relationship into the work integral and solve.
  Working: $W = \int_{V_0}^{2V_0} (aV + b) dV = \left[ \frac{1}{2}aV^2 + bV ight]_{V_0}^{2V_0} = \left(\frac{1}{2}a(2V_0)^2 + b(2V_0)ight) - \left(\frac{1}{2}aV_0^2 + bV_0ight) = 2aV_0^2 + 2bV_0 - \frac{1}{2}aV_0^2 - bV_0 = \frac{3}{2}aV_0^2 + bV_0$
  Key Point: Remember the rules of integration and definite integrals.
 3. State the final answer.
  Result: The work done by the gas is $\frac{3}{2}aV_0^2 + bV_0$.
  Verification: Check the units of the answer.  The units should be Joules (or equivalent energy units).
 ## OPTION ANALYSIS
A :x: Incorrect integration of the pressure-volume relationship.
B ✓: Correct integration and evaluation of the definite integral.
C :x: Incorrect evaluation of the definite integral.
D :x: Incorrect integration and evaluation of the definite integral.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly integrating the expression for pressure: Forgetting to apply the power rule of integration correctly.
  2.Errors in evaluating the definite integral: Mistakes in substituting the limits of integration.
 **Quick Tips**:
  1.Always carefully check your integration steps and substitution of limits.
  2.Double-check units to verify the correctness of your answer.

ABDUL JILANI BASHA · Answer

$aV_0^2 + bV_0$

ABDUL JILANI BASHA · Answer

The correct Option is "$\frac{3}{2}aV_0^2 + bV_0$"

ABDUL JILANI BASHA · Answer

$2aV_0^2 + 2bV_0$

ABDUL JILANI BASHA · Answer

$\frac{1}{2}aV_0^2 + \frac{1}{2}bV_0$

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