Question

At constant pressure, the ratio of increase in volume of an ideal gas per degree rise in kelvin temperature to its original volume is:
${\text{A}}{\text{. }}{T^2} \\\ {\text{B}}{\text{. }}\dfrac{1}{T} \\\ {\text{C}}{\text{. }}{T^3} \\\ {\text{D}}{\text{. }}T \\\$

Answer 1

Hint: We need to calculate the ratio of the change in volume to the original volume of the ideal gas. We know the ideal gas equation which can be used to calculate this ratio at constant pressure.

Formula used:
The ideal gas equation is given as
$PV = nRT$
where P is used to represent the pressure of the ideal gas;
V is used to represent the volume of the ideal gas
n is used to represent the no, of moles of the ideal gas
T is used to represent the temperature of the gas
R is called the gas constant whose value is given as 8.314 J/mol.K

Detailed step by step solution:
We are given an ideal gas which is at constant pressure. Its temperature is increasing, leading to an increase in volume of the gas. For an ideal gas, we can write the ideal gas equation as follows:
$PV = nRT \\\ \Rightarrow V = \left( {\dfrac{{nR}}{P}} \right)T \\\$
The term in the bracket is a constant, therefore , we can write that
$V \propto T$
If ${V_1}$ and ${T_1}$ are initial volume and temperature of the gas and ${V_2}$ and ${T_2}$ final volume and temperature of the gas then we can write the following relation for them.
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}} \\\ or{\text{ }}\dfrac{{{V_2}}}{{{V_1}}} = \dfrac{{{T_2}}}{{{T_1}}} \\\$
Subtracting 1 from both the sides of the equation, we get
$\dfrac{{{V_2}}}{{{V_1}}} - 1 = \dfrac{{{T_2}}}{{{T_1}}} - 1 \\\ \Rightarrow \dfrac{{{V_2} - {V_1}}}{{{V_1}}} = \dfrac{{{T_2} - {T_1}}}{{{T_1}}} \\\$
We are given that there is only 1 degree rise in temperature on the Kelvin scale.
$\therefore {T_2} - {T_1} = 1K$
Using this, we get
$\dfrac{{{V_2} - {V_1}}}{{{V_1}}} = \dfrac{1}{{{T_1}}} \\\ or{\text{ }}\dfrac{{{V_2} - {V_1}}}{{{V_1}}} = \dfrac{1}{T} \\\$
This implies that change in volume per unit original volume at constant pressure is inversely proportional to temperature of the ideal gas. Hence, the correct answer is option B.

Note: The thermodynamic process which takes place at constant pressure is known as an isobaric process. In this type of process, the volume of the gas changes with the change in temperature of the gas.