
For a gas$\dfrac{R}{{{C_V}}} = 0.67$ . This gas is made up of molecules which are
(A) Diatomic
(B) Mixture of diatomic and polyatomic molecules
(C) Monoatomic
(D) Polyatomic
Answer
163.8k+ views
Hint:First start with finding the relation between the gas constant, heat capacity at constant volume and heat capacity at constant pressure in case of the monoatomic gas. Then use the relation and compare it with the relation given in the question and finally get the required answer.
Formula Used:
The Gas constant is given by;
$R = {C_P} - {C_V}$
Complete answer:
Let start with the information given in the question:
Here gas constant is given as R.
Heat capacity at constant volume is given as ${C_V}$ .
Also the value of their ratio is also given as;
$\dfrac{R}{{{C_V}}} = 0.67$
Now,
$R = {C_P} - {C_V}$
Dividing by ${C_V}$ , we get;
$\dfrac{R}{{{C_V}}} = \dfrac{{{C_P} - {C_V}}}{{{C_V}}}$
After solving, we get;
$\dfrac{{{C_P}}}{{{C_V}}} - 1 = 0.67$
So, $\dfrac{{{C_P}}}{{{C_V}}} = 1.67$
We know that: Value of Gamma, $Y = \dfrac{{{C_P}}}{{{C_V}}} = 1.67$is for monoatomic gas.
Hence the correct answer is Option(C).
Note: First try to find the value of gamma then you will get whether the gas is diatomic, monoatomic, polyatomic and so on. The value of gamma is different in all cases. It is 1.67 for monoatomic hence the gas here is monoatomic. It is not the case all the time therefore it depends on the value of gamma, Y.
Formula Used:
The Gas constant is given by;
$R = {C_P} - {C_V}$
Complete answer:
Let start with the information given in the question:
Here gas constant is given as R.
Heat capacity at constant volume is given as ${C_V}$ .
Also the value of their ratio is also given as;
$\dfrac{R}{{{C_V}}} = 0.67$
Now,
$R = {C_P} - {C_V}$
Dividing by ${C_V}$ , we get;
$\dfrac{R}{{{C_V}}} = \dfrac{{{C_P} - {C_V}}}{{{C_V}}}$
After solving, we get;
$\dfrac{{{C_P}}}{{{C_V}}} - 1 = 0.67$
So, $\dfrac{{{C_P}}}{{{C_V}}} = 1.67$
We know that: Value of Gamma, $Y = \dfrac{{{C_P}}}{{{C_V}}} = 1.67$is for monoatomic gas.
Hence the correct answer is Option(C).
Note: First try to find the value of gamma then you will get whether the gas is diatomic, monoatomic, polyatomic and so on. The value of gamma is different in all cases. It is 1.67 for monoatomic hence the gas here is monoatomic. It is not the case all the time therefore it depends on the value of gamma, Y.
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