Question

For an ideal monatomic gas during any process \[{\text{T = kV}}\], the molar heat capacity of the 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 1

Hint:To answer this question, you must recall the concept of molar heat capacity. The amount of heat that is required to raise the temperature of an ideal gas by 1 Kelvin at a constant pressure is known as the molar heat capacity of the gas. We can find the molar heat capacity using the first law of thermodynamics.
Formula used: Formula used: First Law of Thermodynamics: $\Delta {\text{U}} = {\text{Q}} - {\text{W}}$
Also we know that heat can be written as $Q = nC\Delta T$
This gives us molar heat capacity as $C = \dfrac{Q}{{n\Delta T}}$

Complete step by step answer:
In the question, the process occurring in the system is given as \[{\text{T = kV}}\].
Since, the temperature is varying directly with the volume of the gas, thus, we can conclude that pressure of the system is constant. At constant pressure, the work done during a process can be given as, ${\text{W}} = {{P\Delta V}} = {{nR\Delta T}}$
The heat change at constant pressure is given as ${Q_P} = n{C_P}\Delta T$
We know that, the internal energy at constant pressure is given as $\Delta U = \dfrac{3}{2}nR\Delta T$
Substituting these values in the first law expression, we get,
$n{C_P}\Delta T = \dfrac{3}{2}nR\Delta T + {{nR\Delta T}}$
$ \Rightarrow {C_P} = \dfrac{5}{2}R$

Thus, the correct answer is A.
Note:
The first law of thermodynamics is an adapted form of the law of conservation of energy which has been modified so as to be applied in thermodynamic processes. It distinguishes two types of transfer of energy, as the heat transferred and the thermodynamic work done, during a process and relates them to the internal energy of the system which is a function of a system's state.