Solveeit Logo

Question

Question: The Poisson’s ratio (\(\gamma \)) for \({O_2}\) is 1.4. Select incorrect option for \({O_2}\) (R = 2...

The Poisson’s ratio (γ\gamma ) for O2{O_2} is 1.4. Select incorrect option for O2{O_2} (R = 2cal/molKcal/molK)
(A) CV=5 cal/molK{C_V} = 5{\text{ }}cal/molK
(B) CV or SV=0.45cal/molK{C_V}{\text{ or }}{{\text{S}}_V} = 0.45cal/molK
(C) CP=Rγγ1{C_P} = \dfrac{{{R_\gamma }}}{{\gamma - 1}}
(D) CV=Rγ1{C_V} = \dfrac{R}{{\gamma - 1}}

Explanation

Solution

Poisson’s ratio is the ratio of heat capacity at constant pressure to the heat capacity at constant volume. This formula can be given as
γ=CPCV\gamma = \dfrac{{{C_P}}}{{{C_V}}}

Complete answer:
We know that the Poisson’s ratio is the ratio of CP{C_P} to CV{C_V}. Thus, we can write for oxygen gas that
γ=CPCV=1.4\gamma = \dfrac{{{C_P}}}{{{C_V}}} = 1.4
We know that CV{C_V} is the heat capacity at constant volume and CP{C_P} is the heat capacity measured at constant pressure.
- We also know that CP{C_P} and CV{C_V} are related by the following equation.
CPCV=R{C_P} - {C_V} = R
So, we can write that CP=R+CV{C_P} = R + {C_V}
Now, if we put the value of CP{C_P} in the formula of Poisson’s ratio, we get
γ=R+CVCV\gamma = \dfrac{{R + {C_V}}}{{{C_V}}}
So, γ=RCV+CVCV\gamma = \dfrac{R}{{{C_V}}} + \dfrac{{{C_V}}}{{{C_V}}}
Thus, we can write that
γ=RCV+1\gamma = \dfrac{R}{{{C_V}}} + 1
So, we obtain that CV=Rγ1 .....(1){C_V} = \dfrac{R}{{\gamma - 1}}{\text{ }}.....{\text{(1)}}
Similarly, we can write that CV=CPR{C_V} = {C_P} - R
putting this value of CV{C_V} in Poisson’s ratio will give
γ=CPCPR\gamma = \dfrac{{{C_P}}}{{{C_P} - R}}
From this, we obtain that CPCpγ=R{C_P} - \dfrac{{{C_p}}}{\gamma } = R
So, we can write that CP=γRγ1{C_P} = \dfrac{{\gamma R}}{{\gamma - 1}}
We are given that Poisson’s ratio for oxygen (O2{O_2}) is given as 1.4 and R is the ideal gas constant.
The value of R is 8.341 J/molKJ/molK and 1.987cal/molKcal/molK.
- We need to give the value of CV{C_V} in cal/molK. So, we will put the value of R in cal/molK.
Now, we can put the value of R and γ\gamma in equation (1) to obtain
CV=1.9871.41=1.9870.4=4.9675cal/molK{C_V} = \dfrac{{1.987}}{{1.4 - 1}} = \dfrac{{1.987}}{{0.4}} = 4.9675cal/molK
Thus, we can take the value of CV{C_V} as 5 cal/molK.

Thus, the option that is not correct is (B).

Note:
Note that the value of universal gas constant (R) is 8.341 J/molK. J/molK is the SI unit of R. However, if we want to express the energy in calories, then we can take the unit of R as cal/molK.