
For an ideal monatomic gas during the process T=kV . The molar heat capacity of gas during the process is, (assume vibrational degree of freedom to be active)
A.$\dfrac{5}{2}R$
B.$3R$
C.$\dfrac{7}{2}R$
D.$4R$
Answer
546.3k+ views
Hint: A molecule possesses translational, rotational and vibrational degrees of freedom. Energy of a molecule is divided between its translational, rotational and vibrational degree of freedom.
Complete step by step answer:
Given that,
For an ideal monatomic gas during the process T=kV. This means that temperature is directly proportional to volume of gas.
$T \propto V$
Which in term means that pressure is constant. Therefore the molar heat capacity of gas during the process will be equal to molar heat capacity at constant pressure (${C_p}$).
We have the relation, ${C_p} - {C_v} = R$
Where ${C_v}$ is the molar heat capacity at constant volume and R is gas constant. We can get the value of ${C_v}$ from the degree of freedoms and then we can calculate ${C_p}$ using the above relation.
First let us calculate the number of translational, rotational and vibrational degrees of freedom of the molecule.
Unless otherwise mentioned, the number of translational degrees of freedom of a molecule is three. For a linear molecule the number of rotational degrees of freedom is two. The given molecule is a monatomic gas. So there will be no rotational degree of freedom. Number of vibrational degrees of freedom of a molecule is $3n - t$, where n is the number of atoms in a molecule t is the sum of translational and rotational degrees of freedom.
Hence for the given molecule,
The number of translational degree of freedom $ = 3$
The number of rotational degree of freedom $ = 0$
The number of vibrational degree of freedom = $ = 3 \times 1 - 3 = 0$
Hence total number of degree of freedom $ = 3$
Now,
Molar heat capacity at constant volume, ${C_v} = \dfrac{1}{2}R \times $ total number of degree of freedom
${C_v} = \dfrac{1}{2}R \times 3 = \dfrac{3}{2}R$
Let us substitute this value on ${C_p} - {C_v} = R$.
${C_p} - \dfrac{3}{2}R = R$
${C_p} = R + \dfrac{3}{2}R$
${C_p} = \dfrac{5}{2}R$
Hence the molar heat capacity of gas during the process is $\dfrac{5}{2}R$. Option A is correct.
Note:
The vibrational mode will not be active in all cases. In those situations, we do not consider vibrational degrees of freedom while calculating total degrees of freedom. Here in the question it is given that the vibrational degree of freedom is active.
Complete step by step answer:
Given that,
For an ideal monatomic gas during the process T=kV. This means that temperature is directly proportional to volume of gas.
$T \propto V$
Which in term means that pressure is constant. Therefore the molar heat capacity of gas during the process will be equal to molar heat capacity at constant pressure (${C_p}$).
We have the relation, ${C_p} - {C_v} = R$
Where ${C_v}$ is the molar heat capacity at constant volume and R is gas constant. We can get the value of ${C_v}$ from the degree of freedoms and then we can calculate ${C_p}$ using the above relation.
First let us calculate the number of translational, rotational and vibrational degrees of freedom of the molecule.
Unless otherwise mentioned, the number of translational degrees of freedom of a molecule is three. For a linear molecule the number of rotational degrees of freedom is two. The given molecule is a monatomic gas. So there will be no rotational degree of freedom. Number of vibrational degrees of freedom of a molecule is $3n - t$, where n is the number of atoms in a molecule t is the sum of translational and rotational degrees of freedom.
Hence for the given molecule,
The number of translational degree of freedom $ = 3$
The number of rotational degree of freedom $ = 0$
The number of vibrational degree of freedom = $ = 3 \times 1 - 3 = 0$
Hence total number of degree of freedom $ = 3$
Now,
Molar heat capacity at constant volume, ${C_v} = \dfrac{1}{2}R \times $ total number of degree of freedom
${C_v} = \dfrac{1}{2}R \times 3 = \dfrac{3}{2}R$
Let us substitute this value on ${C_p} - {C_v} = R$.
${C_p} - \dfrac{3}{2}R = R$
${C_p} = R + \dfrac{3}{2}R$
${C_p} = \dfrac{5}{2}R$
Hence the molar heat capacity of gas during the process is $\dfrac{5}{2}R$. Option A is correct.
Note:
The vibrational mode will not be active in all cases. In those situations, we do not consider vibrational degrees of freedom while calculating total degrees of freedom. Here in the question it is given that the vibrational degree of freedom is active.
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