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Question: For a diatomic gas, change in internal energy for unit change in temperature at constant pressure an...

For a diatomic gas, change in internal energy for unit change in temperature at constant pressure and volume is U1{U_1} and U2{U_2} respectively , then U1:U2{U_1}:{U_2} is:
A) 5:35:3
B) 7:57:5
C) 1:11:1
D) 5:75:7

Explanation

Solution

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 (Cp{C_p}): It is the amount of heat required to raise the temperature of 1 mole of gas through 1C{1^\circ }C when pressure is kept constant.
Given heat Q=mCpdTQ = m{C_p}dT
Specific heat at constant volume (Cv{C_v}) : It is the amount of heat to raise the temperature of 1 mole of gas through 1C{1^\circ }C , when volume is kept constant.
Given heat Q=mCvdTQ = m{C_v}dT
Internal energy of a gas molecule =12fKT = \dfrac{1}{2}fKT
Internal energy of 1 mole of gas =12fRT = \dfrac{1}{2}fRT
where ff= degree of freedom
RR = universal gas constant
TT = temperature
Change in internal energy : ΔU=12fRΔT\Delta U = \dfrac{1}{2}fR\Delta T
ΔU=nCpΔT U1=nCpΔT  \Delta U = n{C_p}\Delta T \\\ {U_1} = n{C_p}\Delta T \\\ ………(i) ( at constant pressure)
ΔU=nCvΔT U2=nCvΔT  \Rightarrow \Delta U = n{C_v}\Delta T \\\ \Rightarrow {U_2} = n{C_v}\Delta T \\\ ……..(ii) ( at constant volume)
n=n = number of moles.
From eq (i) :
If n=1n = 1 , then
dU=CvdT Cv=dUdT  \Rightarrow dU = {C_v}dT \\\ \Rightarrow {C_v} = \dfrac{{dU}}{{dT}} \\\
Internal energy of 1 mole ideal gas is given by: dU=12fRdTdU = \dfrac{1}{2}fRdT
dU=12fRdT dUdT=12fR  \Rightarrow dU = \dfrac{1}{2}fRdT \\\ \Rightarrow \dfrac{{dU}}{{dT}} = \dfrac{1}{2}fR \\\
This implies that -Cv=12fR{C_v} = \dfrac{1}{2}fR…….(iii)
From Mayer’ s law we know that-
CpCv=R Cp=R+12fR Cp=(2+f2)R  \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-
U1U2=nCpΔTnCvΔT U1U2=CpCv  \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 : Cp=72R,Cv=52R.  {C_p} = \dfrac{7}{2}R, {C_v} = \dfrac{5}{2}R. \\\
The required ratio is 7:57:5.

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.