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

Hint: 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:
$\Delta {\text{H}} = \Delta {\text{U}} + \Delta {{\text{n}}_{\text{g}}}{\text{RT}}$.