Question

Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)
(A) Cyclic process: ${\text{q = - w}}$
(B) Isothermal process: ${\text{q = - w}}$
(C) Adiabatic process: $\Delta {\text{U = - w}}$
(D) Isochoric process: $\Delta {\text{U = q}}$

Answer 1

The mathematical form of the first law of thermodynamics states that change in the internal energy of a system is equal to the sum of the amount of heat added to the system and the work done on the system.
Thus, it can be expressed as:
$\Delta {\text{U = q + w}}$
Here, q is the heat supplied to the system, w is the work done on the system and $\Delta {\text{U}}$ is the change in the internal energy of the system.

Complete step by step answer:
Heat q and work w are not state functions as their values depend upon the path in which a change is carried out. But, energy is a state function and so the value of $\Delta {\text{U}}$ depends upon the initial and final state.
Whatever maybe the process, $\Delta {\text{U}}$ is always equal to ${\text{q + w}}$ and so ${\text{q + w}}$ is also a state function.
During a cyclic process, the system returns to its initial state. Thus, there is no change in the internal energy of the system which means $\Delta {\text{U = 0}}$ . Then from the first law expression, we have:

$${\text{0 = q + w}} \\\ \Rightarrow {\text{q}} = - {\text{w}} \\\$$

Thus, the heat absorbed by the system in a cyclic process is equal to negative of work done on the system or equal to work done by the system. So option A is correct.
During an isothermal process, the temperature of the system remains constant. When temperature is held constant for ideal gases, the internal energy also remains constant. This means $\Delta {\text{U = 0}}$ . Then from the first law expression, we have:

$${\text{0 = q + w}} \\\ \Rightarrow {\text{q}} = - {\text{w}} \\\$$

Thus, the heat absorbed by the system in a cyclic process is equal to negative of work done on the system or equal to work done by the system. So option B is correct.
In an adiabatic process, there is no entry or exit of heat from the system. This means ${\text{q = 0}}$ . Then from the first law expression, we have:

$$\Delta {\text{U = 0 + w}} \\\ \Rightarrow \Delta {\text{U}} = {\text{w}} \\\$$

So option C is incorrect as it does not correctly represent the first law of thermodynamics for adiabatic processes.
In an isochoric process, the volume of the system remains constant or $\Delta {\text{V = 0}}$ .
Since only expansion work is involved in the change that is the work done by the system is only pressure-volume work, then:
${\text{w}} = - {\text{P}}\Delta {\text{V}}$
Here, P is the pressure and $\Delta {\text{V}}$ is the volume change.
So, for isochoric process,

$$\Delta {\text{U = q - P}}\Delta {\text{V}} \\\ \Rightarrow \Delta {\text{U = q - }}0 \\\ \Rightarrow \Delta {\text{U = q}} \\\$$

So, option D is correct.

Note:
The relationship between the change in enthalpy $\Delta {\text{H}}$ and change in internal energy of a system is given by the relation $\Delta {\text{H = }}\Delta {\text{U}} + {\text{P}}\Delta {\text{V}}$ .
If the difference between the number of moles of gaseous products and reactants is $\Delta {{\text{n}}_{\text{g}}}$ , and R and T is the gas constant and temperature respectively, then ${\text{P}}\Delta {\text{V = }}\Delta {{\text{n}}_{\text{g}}}{\text{RT}}$ . Then we get:
$\Delta {\text{H}} = \Delta {\text{U}} + \Delta {{\text{n}}_{\text{g}}}{\text{RT}}$.