Question 96: Thermodynamics & KTG

Question

An ideal gas at pressure $P$ and temperature $T$ is expanding such that $P T^3=$ constant. The coefficient of volume expansion of the gas is $\_\_\_\_$ .

Accepted Answer

The coefficient of volume expansion $(\gamma)$ is defined as the fractional change in volume per unit change in temperature at constant pressure :
$$ \gamma=\frac{1}{\mathrm{~V}}\left(\frac{\mathrm{dV}}{\mathrm{dT}}ight) $$
The relation between pressure and temperature is,
$$ \mathrm{PT}^3=\mathrm{C} 	ext { (where } \mathrm{C} 	ext { is a constant) } $$
For an ideal gas, $\mathrm{PV}=\mathrm{nRT} \Rightarrow \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$
Substitute P into the given process equation :
$$ \left(\frac{\mathrm{nRT}}{\mathrm{~V}}ight) \mathrm{T}^3=\mathrm{C} $$
$$\Rightarrow $$ $\frac{\mathrm{nRT}^4}{\mathrm{~V}}=\mathrm{C}$
$$ \Rightarrow \mathrm{V}=\left(\frac{\mathrm{nR}}{\mathrm{C}}ight) \mathrm{T}^4 $$
Let $\mathrm{k}=\frac{\mathrm{nR}}{\mathrm{C}}$ (another constant), then:
$$ \mathrm{V}=\mathrm{kT}^4 $$
On differentiating $V$ with respect to $T$ :
$$ \frac{\mathrm{dV}}{\mathrm{dT}}=\frac{\mathrm{d}}{\mathrm{dT}}\left(\mathrm{kT}^4ight)=4 \mathrm{kT}^3 $$
Substituting the expressions for V and $\frac{\mathrm{dV}}{\mathrm{dT}}$ into the definition of $\gamma$ :
$$ \gamma=\frac{1}{\mathrm{~V}}\left(\frac{\mathrm{dV}}{\mathrm{dT}}ight) $$
$$\Rightarrow $$ $\gamma=\frac{1}{\mathrm{kT}^4}\left(4 \mathrm{kT}^3ight)$
$$\Rightarrow $$ $\gamma=\frac{4 \mathrm{kT}^3}{\mathrm{kT}^4}$
$$\Rightarrow $$ $$ \gamma=\frac{4}{\mathrm{~T}} $$
Therefore, the coefficient of volume expansion of the gas is $\frac{4}{T}$.
Hence, the correct option is (C).

Answer

$\frac{2}{T}$

Answer

$\frac{1}{T}$

Answer

$\frac{4}{T}$

Answer

$\frac{3}{T}$

Thermodynamics & KTG

An ideal gas at pressure P and temperature T is expanding such that P T^3= constant. The coefficient of volume expansion of the gas is \_\_\_\_ .

Choose the correct answer: