Heat And Thermodynamics Ques 147
- One mole of a monoatomic ideal gas undergoes four thermodynamic processes as shown schematically in the $p V$-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-l with the corresponding statements in List-II.
(2018 Adv.)
| List-I | List-II | ||
|---|---|---|---|
| P. | In process I | 1. | Work done by the gas is zero |
| Q. | In process II | 2. | Temperature of the gas remains unchanged |
| R. | In process III | 3. | No heat is exchanged between the gas and its surroundings |
| S. | In process IV | 4. | Work done by the gas is $6 p _0 V _0$ |
(a) $P \rightarrow 4$; $Q \rightarrow 3$; $R \rightarrow 1$; $S \rightarrow 2$
(b) $P \rightarrow 1 ; Q \rightarrow 3 ; R \rightarrow 2 ; S \rightarrow 4$
(c) $P \rightarrow 3$; $Q \rightarrow 4$; $R \rightarrow 1 ; S \rightarrow 2$
(d) $P \rightarrow 3$; $Q \rightarrow 4 ; R \rightarrow 2 ; S \rightarrow 1$
Show Answer
Answer:
Correct Answer: 147.(c)
Solution:
Formula:
- $\left(\frac{d p}{d V}\right) _{\text {adiabatic }}=\gamma\left(\frac{d p}{d V}\right) _{\text {isothermal }}$
List-I
(P) Process I $\quad \Rightarrow$ Adiabatic $\Rightarrow Q=0$
(Q) Process II $\Rightarrow$ Isobaric
$\therefore \quad W=p \Delta V=3 p _0\left[3 V _0-V _0\right]=6 p _0 V _0$
(R) Process III $\Rightarrow$ Isochoric $\Rightarrow W=0$
(S) Process (IV) $\Rightarrow$ Isothermal $\Rightarrow$ Temperature $=$ Constant
BETA
