Question

Which of the following is not correct for an ideal gas as per the first law of thermodynamics?

• A. Adiabatic, $∆\mathrm{U}=-\mathrm{w}$
• B. Isothermal, $\mathrm{q}=-\mathrm{w}$
• C. Cyclic, $\mathrm{q}=-\mathrm{w}$
• D. Isochoric, $∆\mathrm{U}=\mathrm{q}$

Answer 1

The correct option is A

Adiabatic, $∆\mathrm{U}=-\mathrm{w}$

An explanation for the correct answer:

(a) Adiabatic

1. The energy ($∆\mathrm{Q}$) supplied to the system goes into the part to increase the internal energy ($∆\mathrm{U}$) of the system and the rest in the work ($∆\mathrm{W}$) for the environment.
2. The general principle for the conservation of energy is: $∆\mathrm{Q}=∆\mathrm{U}+∆\mathrm{W}$
3. Adiabatic process: The system is insulated from the surroundings and the absorption or release of the heat is zero.
4. The general expression for the adiabatic process of the ideal gas is: ${\mathrm{pv}}^{\mathrm{\gamma }}=\mathrm{constant}$, where $\mathrm{\gamma }=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{\vartheta }}}$ (The ratio of the specific heats at constant pressure and temperature.

An explanation for the incorrect answer:

(b) Isothermal process:

1. There is no change in the temperature and the internal energy depends on the temperature. So, there is no change in the internal energy of an ideal gas. $∆\mathrm{U}={\mathrm{mC}}_{\mathrm{p}}∆\mathrm{T}\mathrm{and}∆\mathrm{U}={\mathrm{mC}}_{\mathrm{\vartheta }}∆\mathrm{T}$
2. The heat supplied to the gas is equal to the work done by the gas. $\mathrm{Q}=\mathrm{W}$

(c) Cyclic process:

1. The system returns to its original state since the internal energy of the system is zero.
2. The total heat absorbed is equal to the work done by the system.$∆\mathrm{U}=0$

(d) Isochoric process:

1. In this process, the volume is constant so that there is no work done on or by the gas.
2. The heat absorbed by the gas goes entirely to change its internal energy and its temperature.$\mathrm{Q}=∆\mathrm{U}$

Therefore, option (a) is correct.