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In Mayer’s relation${{{C}}_{{p}}}{{ - }}{{{C}}_{{v}}}{{ = R}}$‘R’ stands for:A.Translational kinetic energy of 1 mol of gas B.Rotational kinetic energy of 1 mol gasC.Vibrational kinetic energy of 1 mol gasD.Work done to increase the temperature of 1 mol gas by one degree

Last updated date: 25th Jun 2024
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Hint:Specific heat is the heat required to raise the temperature of the unit mass of a gas. ${{{C}}_{{v}}}$ is the specific heat capacity at constant volume. ${{{C}}_{{p}}}$ is the specific heat capacity at constant pressure. According to Mayer, the specific heat capacity at constant pressure is greater than specific heat at constant volume.

Complete step by step answer:
Let us take 1 mole of an ideal gas in a container and a piston.
The piston is fixed tightly on the top of the container and the ideal gas is heated.
We can understand that as the piston is fixed, the ideal gas in the container would not be able to expand.
So we can say that the ideal gas is heated at constant volume as the gas cannot be expanded because the piston is not movable.
At constant volume, heat ${{H = n}}{{{C}}_{{v}}}{{\Delta T}}$ and for a small heat change ${{dH = }}{{{C}}_{{v}}}{{dT}}$ because the number of mole of ideal gas taken is 1.
We know that work can be expressed in terms of pressure and change in volume
${{dW = PdV}}$ . Since as discussed above there is no volume change, the work done is zero at constant volume.
${{H = U + PV}}$ according to the first law of thermodynamics. So we can write as ${{dH = dU + dW}}$ where work is a function of pressure and temperature and at constant volume, it is zero.
So we can say that ${{dH = dU}}$
$\Rightarrow {{ }}{{{C}}_v}{{dT = dU}}$
Now let us take a movable piston. So, on heating temperature raise to dT at constant pressure. We can say that ${{H = }}{{{C}}_{{p}}}{{\Delta T}}$ and as we discussed above ${{dW = PdV}}$ and ${{dH = dU + dW}}$.
So it can be rewritten as ${{ }}{{{C}}_{{p}}}{{dT = dU + PdV}}$ and dU is dependent on the temperature change.
${{ }}{{{C}}_{{p}}}{{dT = dU + PdV}}$
$\Rightarrow {{ }}{{{C}}_{{p}}}{{dT = }}{{{C}}_{{v}}}{{dT + PdV}} \to eq1$
We know the PdV is the work done.
From the ideal gas equation, ${{{C}}_{{p}}}{{dT = }}{{{C}}_{{v}}}{{dT + PdV}}$and change in volume and temperature of 1 mole can be ${{PdV = RdT}}$ . We know that ${{PV = nRT}}$
Substituting it in equation 1, we get $\Rightarrow {{ }}{{{C}}_{{p}}}{{dT = }}{{{C}}_{{v}}}{{dT + RdT}}$
On canceling dT we get ${{ }}{{{C}}_{{p}}}{{ = }}{{{C}}_{{v}}}{{ + R}}$
Therefore, we can say that R stands for PdV that is the work done to increase the temperature of 1 mole of gas by one degree.
The correct option is (D).

Note:
When we measure the temperature of a gas, the average translational kinetic energy of its molecules is measured. In a diatomic molecule, for each atom, there are axes of rotation. For a diatomic gas, there are 3 degrees of freedom, three translational, and 2 rotational. R is the gas constant which is in Mayer’s relation considered as the work is done.