Question

In an isobaric process, $\Delta Q=\dfrac{K\gamma }{\gamma -1}$ where $\gamma =\dfrac{{{C}_{p}}}{{{C}_{v}}}$. What is K?
$\text{A}\text{. Pressure}$
$\text{B}\text{. Volume}$
$\text{C}\text{. }\Delta U$
$\text{D}\text{. }\Delta W$

Answer 1

In thermodynamics, isobaric process is a process in which pressure of the system remains constant throughout the process.
First law of thermodynamics will be helpful in solving this problem. According to this law, when heat is added to a system it is only used to change its internal energy and do some work.

Formula used: For an isobaric process, Heat $\Delta Q=n{{C}_{p}}\Delta T$ and work done on the system $\Delta W=nR\Delta T$

Complete step by step answer:
First law of thermodynamics states, “the internal energy of a system has to be equal to the work that is being done on the system, plus or minus the heat that flows in or out of the system respectively and any other work that is done on the system.”
If $\Delta Q$ is the heat added to system, $\Delta U$ be the change in its internal energy and, $\Delta W$ is the work done on the system then according to first law of thermodynamics
$\Delta Q=\Delta U+\Delta W$
This law is based on conservation of energy, for heat being a form of energy can neither be created nor destroyed.
A thermodynamic process is the process when the values of thermodynamic variables associated with a system change from one equilibrium to another.
In an isobaric process, pressure does not change throughout the process.
For an isobaric process, $\Delta U=n{{C}_{v}}\Delta T$and $\Delta W=nR\Delta T$
Therefore according to first law,
$\Delta Q=n{{C}_{v}}\Delta T+nR\Delta T$
Since ${{C}_{p}}={{C}_{v}}+R$
$\Delta Q=n{{C}_{p}}\Delta T$ …..(1)
Where ${{C}_{p}}$and ${{C}_{v}}$ are molar heat capacity at constant pressure and volume respectively. $R$ is the universal gas constant and $n$is the number of moles of gas.
According to the question, $\Delta Q=\dfrac{K\gamma }{\gamma -1}$
Let us simplify this equation by substituting $\gamma =\dfrac{{{C}_{p}}}{{{C}_{v}}}$ and the equation it with equation (1).
$\Delta Q=\dfrac{K\dfrac{{{C}_{p}}}{{{C}_{v}}}}{\dfrac{{{C}_{p}}}{{{C}_{v}}}-1}=\dfrac{K{{C}_{p}}}{{{C}_{p}}-{{C}_{v}}}$
$\Delta Q=\dfrac{K{{C}_{p}}}{R}$ (because ${{C}_{p}}={{C}_{v}}+R$)
$\Rightarrow \Delta Q=\dfrac{K{{C}_{p}}}{R}=n{{C}_{p}}\Delta T$
On solving, we get
$K=nR\Delta T=\Delta W$

So, the correct answer is “Option D”.

Note: Heat is defined as energy in transit as a result of temperature difference.
Work done, in the thermodynamics process, is defined as a product of pressure on the system and change in its volume. In an isobaric process, for an ideal gas, if P and $V$ are pressure and the volume of the system respectively, then
$\Delta W=P\Delta V=nR\Delta T$