The pressure (P) and temperature (T) relationship of an ideal gas obeys the equation PT^2 = constant. The volume expansion coefficient of the gas will be:

Given: For the ideal gas, PT^2 = constant. Find: The volume expansion coefficient. From the solution, use dVV = dTT For a constant PT^2, it states: dVdT = (C) 3T^2 So, the volume expansion coefficient is written as dVV = 3T^2 dT Hence, the extracted solution concludes that the correct option is B, i.e. 3T^2. The solution's shows a disagreement between the listed correct answer and the solution, and the solution is taken.

The pressure (P) and temperature (T) relationship of an ideal gas… — Solution | JEE Physics

MCQMediumJEE 2023Equation of State of Ideal Gas

The pressure ( $P$ ) and temperature ( $T$ ) relationship of an ideal gas obeys the equation $PT^2 = \text{constant}$ . The volume expansion coefficient of the gas will be:

A
$3T^2$
B
$\frac{3}{T^2}$
C
$\frac{3}{T^3}$
D
$\frac{3}{T}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: For the ideal gas, $PT^2 = \text{constant}$ .

Find: The volume expansion coefficient.

From the solution, use

\frac{dV}{V} = \alpha \frac{dT}{T}

For a constant $PT^2$ , it states:

\frac{dV}{dT} = \left(C\right)\frac{3}{T^2}

So, the volume expansion coefficient is written as

\frac{dV}{V} = \frac{3}{T^2} \cdot dT

Hence, the extracted solution concludes that the correct option is B, i.e. $\frac{3}{T^2}$ . The solution's shows a disagreement between the listed correct answer and the solution, and the solution is taken.

Common mistakes

Using the answer key key without checking the worked solution. Here the solution's contains a mismatch, so the solution working and conclusion must be treated as primary.
Confusing the definition of volume expansion coefficient with a direct derivative like $\frac{dV}{dT}$ . The coefficient is defined through a fractional change in volume, so the correct relation must involve $\frac{dV}{V}$ .
Ignoring the given constraint $PT^2 = \text{constant}$ while applying the ideal gas relation. The condition changes how $P$ varies with $T$ , so one cannot assume pressure is constant.

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