The Question is from neet-chemistry chemical-thermodynamics , and it was created by the author Mr. S M HUSSAIN

Question

An ideal gas undergoes a process in which its pressure P and volume V are related by $PV^n = constant$, where n is a constant.  If the gas expands adiabatically ($n = \gamma$, where $\gamma = C_P/C_V$), which of the following statements is TRUE regarding the work done by the gas?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Thermodynamic work done by an ideal gas
 **Related Topics**: Adiabatic processes, specific heat capacities, ideal gas law
 **Prerequisites**: Understanding of thermodynamic processes, work-energy theorem, and ideal gas equations.
## SOLUTION PATHWAY
**Given Information**:
  $PV^n = constant$
  Adiabatic process: $n = \gamma = C_P/C_V$
  Gas expands
**Method & Steps**:
 1. The work done by an ideal gas during a process is given by:
  $W = \int_{V_i}^{V_f} P dV$
  Reasoning: We need to express P in terms of V using the given relation $PV^n = constant = PV_i^n = PV_f^n$ for an adiabatic process. 
  Formula/Rule: $P = \frac{constant}{V^n} = \frac{P_i V_i^n}{V^n}$
 2. Substitute the expression for P into the work integral:
  $W = \int_{V_i}^{V_f} \frac{P_i V_i^n}{V^n} dV = P_i V_i^n \int_{V_i}^{V_f} V^{-n} dV$
  Working: $W = P_i V_i^n \left[ \frac{V^{-n+1}}{-n+1} ight]_{V_i}^{V_f} = \frac{P_i V_i^n}{1-n} (V_f^{1-n} - V_i^{1-n})$
  Since $PV^n = constant$, $P_i V_i^n = P_f V_f^n$. Using the ideal gas law ($PV = nRT$), we can substitute $P_i V_i = nRT_i$ and $P_f V_f = nRT_f$.
  $W = \frac{nR}{1 - \gamma} (T_f - T_i)$ 
  Key Point: The work done depends on both initial and final temperatures.
 3. Final step:  The work done is explicitly dependent on both $T_i$ and $T_f$ for an adiabatic process.
  Result: The work done depends on both the initial and final temperatures.
  Verification: The derived formula for work confirms the dependence on both initial and final temperatures.
 ## OPTION ANALYSIS
A ✓: The work done depends on both the initial and final temperatures. (Correct, as shown in the solution)
B :x: The work done is zero. (Incorrect, work is done during expansion)
C :x: The work done depends only on the initial temperature. (Incorrect, final temperature also plays a role)
D :x: The work done depends only on the final temperature. (Incorrect, initial temperature also plays a role)
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly assuming the work done is zero for an adiabatic process. (Remember, adiabatic doesn't mean isothermal.)
  2.Forgetting to use the correct expression for pressure in terms of volume for the given adiabatic process. 
 **Quick Tips**:
  1.Remember the general formula for work done by a gas: $W = \int_{V_i}^{V_f} P dV$
  2.For adiabatic processes, use $PV^\gamma = constant$ and ideal gas law to relate work to temperatures.

Mr. S M HUSSAIN · Answer

The correct Option is "The work done depends on both the initial and final temperatures."

Mr. S M HUSSAIN · Answer

The work done is zero.

Mr. S M HUSSAIN · Answer

The work done depends only on the initial temperature.

Mr. S M HUSSAIN · Answer

The work done depends only on the final temperature.

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