Question

For a diatomic gas, change in internal energy for unit change in temperature at constant pressure and volume is ${U_1}$ and ${U_2}$ respectively , then ${U_1}:{U_2}$ is:
A) $5:3$
B) $7:5$
C) $1:1$
D) $5:7$

Answer 1

Hint: Formulate the internal energy of the diatomic gas . And then calculate the change in internal energy for both pressure and volume . Put the values of change for pressure and volume and calculate the required ratio.

Complete step by step solution:
Internal energy is actually the sum of all the microscopic energy of a system. It depends on the molecular structure and the degree of molecular activity . It is the sum of the kinetic energy and potential energy of the molecules.
Specific Heats of a gas :
Specific heat at constant pressure (${C_p}$): It is the amount of heat required to raise the temperature of 1 mole of gas through ${1^\circ }C$ when pressure is kept constant.
Given heat $Q = m{C_p}dT$
Specific heat at constant volume (${C_v}$) : It is the amount of heat to raise the temperature of 1 mole of gas through ${1^\circ }C$ , when volume is kept constant.
Given heat $Q = m{C_v}dT$
Internal energy of a gas molecule $ = \dfrac{1}{2}fKT$
Internal energy of 1 mole of gas $ = \dfrac{1}{2}fRT$
where $f$= degree of freedom
$R$ = universal gas constant
$T$ = temperature
Change in internal energy : $\Delta U = \dfrac{1}{2}fR\Delta T$
$
  \Delta U = n{C_p}\Delta T \\
  {U_1} = n{C_p}\Delta T \\
 $ ………(i) ( at constant pressure)
$
\Rightarrow \Delta U = n{C_v}\Delta T \\
\Rightarrow {U_2} = n{C_v}\Delta T \\
 $ ……..(ii) ( at constant volume)
$n = $number of moles.
From eq (i) :
If $n = 1$ , then
$
\Rightarrow dU = {C_v}dT \\
\Rightarrow {C_v} = \dfrac{{dU}}{{dT}} \\
 $
Internal energy of 1 mole ideal gas is given by: $dU = \dfrac{1}{2}fRdT$
$
\Rightarrow dU = \dfrac{1}{2}fRdT \\
\Rightarrow \dfrac{{dU}}{{dT}} = \dfrac{1}{2}fR \\
 $
This implies that -${C_v} = \dfrac{1}{2}fR$…….(iii)
From Mayer’ s law we know that-
$
\Rightarrow {C_p} - {C_v} = R \\
\Rightarrow {C_p} = R + \dfrac{1}{2}fR \\
\Rightarrow {C_p} = (\dfrac{{2 + f}}{2})R \\
 $...........(iv)
 The ratio of both the internal energies is given by-
$
\Rightarrow \dfrac{{{U_1}}}{{{U_2}}} = \dfrac{{n{C_p}\Delta T}}{{n{C_v}\Delta T}} \\
 \Rightarrow \dfrac{{{U_1}}}{{{U_2}}} = \dfrac{{{C_p}}}{{{C_v}}} \\
 $
For diatomic gas : $
  {C_p} = \dfrac{7}{2}R, {C_v} = \dfrac{5}{2}R. \\
 $
The required ratio is $7:5$.

Hence, the correct option is B.

Note: If the temperature is kept constant, the values of specific heats of pressure and volume will be the same. Internal energy will change only if there is a change in temperature.