Question

For perfect gas, the ratio of volume coefficient of expansion to pressure coefficient is
A. Equal to one
B. Less than one
C. More than one
D. An imaginary quantity

Answer 1

Express the coefficient of volume expansion using the linear equation of volume expansion. Use an ideal equation and differentiate it with respect to temperature keeping pressure constant to determine the coefficient of volume expansion. Again differentiate the ideal gas equation with respect to temperature keeping volume constant to determine the coefficient of pressure expansion.

Formula used:
$\Delta V = V\beta \Delta T$
Here, $\beta$ is known as a coefficient of volume expansion, V is the original volume of the material and $\Delta T$ is the change in temperature.

Complete step by step answer:
When we heat a material, it expands in all directions. The volume expansion of the material is given as,
$\Delta V = V\beta \Delta T$
Here, $\beta$ is known as a coefficient of volume expansion, V is the original volume of the material and $\Delta T$ is the change in temperature.

We can express the rate of volume expansion with respect to temperature as follows,
$\dfrac{{\partial V}}{{\partial T}} = \beta V$
$\Rightarrow \beta = \dfrac{1}{V}\dfrac{{\partial V}}{{\partial T}}$
We first consider the coefficient of volume expansion at constant pressure as follows,
${\beta _V} = \dfrac{1}{V}{\left( {\dfrac{{\partial V}}{{\partial T}}} \right)_P}$ …… (1)
We know the ideal gas equation, $PV = nRT$, where, n is the number of moles, R is gas constant and T is the temperature.

We can write the above equation as follows,
$V = \dfrac{{nRT}}{P}$
$\Rightarrow V = \left( T \right)\left( {\dfrac{{nR}}{P}} \right)$

Now, we take the natural logarithm of the above equation.
$\ln V = \ln \left( T \right) + \ln \left( {\dfrac{{nR}}{P}} \right)$
Now, we have to differentiate the above equation with respect to T. Therefore,
$\Rightarrow \dfrac{1}{V}\dfrac{{\partial V}}{{\partial T}} = \dfrac{1}{T} + 0$
$\Rightarrow \dfrac{1}{V}\dfrac{{\partial V}}{{\partial T}} = \dfrac{1}{T}$

Using equation (1), we can write the above equation as,
${\beta _V} = \dfrac{1}{T}$ …… (2)
We can write the ideal gas equation as follows,
$P = \dfrac{{nRT}}{V}$
$\Rightarrow P = \left( T \right)\left( {\dfrac{{nR}}{V}} \right)$

Now, we take the natural logarithm of the above equation.
$\ln P = \ln \left( T \right) + \ln \left( {\dfrac{{nR}}{V}} \right)$
Now, we have to differentiate the above equation with respect to T keeping the volume constant. Therefore,
$\Rightarrow \dfrac{1}{P}\dfrac{{\partial P}}{{\partial T}} = \dfrac{1}{T} + 0$
$\Rightarrow \dfrac{1}{P}\dfrac{{\partial P}}{{\partial T}} = \dfrac{1}{T}$

We know that the term $\dfrac{1}{P}\dfrac{{\partial P}}{{\partial T}}$ is known as pressure coefficient of expansion ${\gamma _P}$. Therefore,
${\gamma _P} = \dfrac{1}{T}$ …… (3)
We divide equation (2) by equation (3).
$\dfrac{{{\beta _V}}}{{{\gamma _P}}} = \dfrac{{\dfrac{1}{T}}}{{\dfrac{1}{T}}}$
$\therefore \dfrac{{{\beta _V}}}{{{\gamma _P}}} = 1$
Therefore, the ratio of volume coefficient of expansion and pressure coefficient of expansion is equal to one.

So, the correct answer is option A.

Note: We have the rate of volume expansion with respect to temperature $\dfrac{{\partial V}}{{\partial T}} = \beta V$. In this equation we have taken the partial derivative of the volume as it is not only a function of volume but also the function of temperature. We can take the linear derivative, if quantity is a function of only one variable.