Question

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 :

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

Answer 1

The Correct Option is A

To solve this problem, we need to calculate the volume expansion coefficient of the gas, given the relationship between pressure \((P)\) and temperature \((T)\) for an ideal gas, expressed as \(PT^2 = \text{constant}\).

1. We start from the ideal gas law: \(PV = nRT\). This relates the pressure \((P)\), volume \((V)\), and temperature \((T)\) of an ideal gas.
2. According to the problem, \(PT^2 = \text{constant}\). Let's refer this constant to be \(k\).
3. Substitute for \(P\) using the given equation: \(P = \frac{k}{T^2}\).
4. According to the ideal gas equation, substitute \(P\): \(\frac{k}{T^2} V = nRT\).
5. Rearrange to express the volume \((V)\): \(V = nR \frac{T^3}{k}\).
6. Now, the volume is proportional to \(T^3\), which implies \(V \propto T^3\).
7. The volume expansion coefficient \((\beta)\) is defined as: \(\beta = \frac{1}{V} \frac{dV}{dT}\).
8. Substitute \(V \propto T^3\), we obtain: \(\beta = \frac{d (T^3)}{T^3 \, dT} = \frac{3T^2}{T^3} = \frac{3}{T} \cdot T^3 = 3T^2\).
9. Therefore, the volume expansion coefficient of the gas is \(3T^2\), which matches option $3 T^2$.

In summary, the correct answer is $3 T^2$, as it corresponds to the volume expansion coefficient given the specified pressure-temperature relationship.