Isobaric Process MCQ - Practice Questions with Answers

Isobaric Process- A Thermodynamic process in which pressure remains constant is known as the isobaric process.

In this process, V and T change keeping P constant. I.e Charle’s law is obeyed in this process

   Key points in the Isobaric Process-

• Its Equation of state is given as $\frac{V}{T}=$ constant

$$\text{ І.e }\frac{V_1}{T_1}=\frac{V_2}{T_2}=\text{ constant }$$

• - P-V Indicator diagram for an isobaric process. Its PV graph has slope $=0$ (i.e $\frac{dP}{dV}=0$,

The above Graph represents an isobaric expansion.

 The P-V diagram for this process is a line parallel to the volume line.

- Specific heat of gas during the isobaric process is given by

$$C_P=(\frac{f}{2}+1)R$$

- The bulk modulus of elasticity during the isobaric process is given by

$$K=\frac{\mathrm{\Delta }P}{-\mathrm{\Delta }V/V}=0$$

- Work done in the isobaric process-

$$\mathrm{\Delta }W={\int }_{V_i}^{V_f}PdV=P{\int }_{V_i}^{V_f}dV=P[V_f-V_i]$$

Or we can write $\mathrm{\Delta }W=P(V_f-V_f)=nR[T_f-T_t]=nR\mathrm{\Delta }T$
- Internal energy in an isobaric process

$$\mathrm{\Delta }U=n{C}_V\mathrm{\Delta }T=n\frac{R}{(\gamma -1)}\mathrm{\Delta }T$$
 

- Heat in an isobaric process

From FLTD $\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$

$$\begin{array}{rl}& \begin{array}{rl}\mathrm{\Delta }Q& =n\frac{R}{(\gamma -1)}\mathrm{\Delta }T+nR\mathrm{\Delta }T=nR\mathrm{\Delta }T[\frac{1}{\gamma -1}+1]\\ \Rightarrow \mathrm{\Delta }Q& =nR\mathrm{\Delta }T\frac{\gamma }{\gamma -1}=n\left(\frac{\gamma }{\gamma -1}\right)R\mathrm{\Delta }T=n{C}_p\mathrm{\Delta }T\end{array}\\ & \text{ So }\mathrm{\Delta }Q=n{C}_p\mathrm{\Delta }T\end{array}$$

- Examples of the isobaric process-
1. Conversion of water into vapor phase (boiling process)

From the first law of thermodynamics

$$\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W=\mathrm{\Delta }{U}_K+\mathrm{\Delta }{U}_P+\mathrm{\Delta }W$$

since $\mathrm{\Delta }{U}_K=0$ [as there is no change in temperature] and using, $\mathrm{\Delta }Q=mL$

$$\begin{array}{rl}\mathrm{\Delta }Q& =\mathrm{\Delta }{U}_P+P[V_f-V_t]\\ \mathrm{\Delta }{U}_P& =\mathrm{\Delta }Q-P[V_f-V_t]\\ \mathrm{\Delta }{U}_P& =mL-P[V_f-V_i]\end{array}$$

2. Conversion of ice into water

From FLOT $\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$ and using, $\mathrm{\Delta }Q=mL$
we get $mL=\mathrm{\Delta }{U}_P+\mathrm{\Delta }{U}_K+\mathrm{\Delta }W$

$$mL=\mathrm{\Delta }{U}_P+\mathrm{\Delta }{U}_K+P(V_f-V_i)$$

since $\mathrm{\Delta }{U}_K=0$ [as there is no change in temperature]
and $\mathrm{\Delta }W=0$ [As $V_f-V_{i\text{ is negligible, I.e, when ice converts into water then changes }}$ in volume, is negligible]
Hence $\mathrm{\Delta }{U}_P=mL$