Question

One mole of a monoatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV- 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 - I with the corresponding statements in List - II.

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 $6P_0{V}_0$

• A. $P\to 4;Q\to 3;R\to 1;S\to 2$
• B. $P\to 1;Q\to 3;R\to 2;S\to 4$
• C. $P\to 3;Q\to 4;R\to 1;S\to 2$
• D. $P\to 3;Q\to 4;R\to 2;S\to 1$

Answer 1

The correct option is C $P\to 3;Q\to 4;R\to 1;S\to 2$

$\text{Solution}:$

Process - I is an adiabatic process

$\Delta Q=\Delta U+W\Delta Q=0$

$W=-\Delta U$

Volume of gas is decreasing $\Rightarrow$ W < 0

$\Delta U>0$

$\Rightarrow$ Temperatuer of gas increases.

$\Rightarrow$ No heat is exchanged between the gas and surrounding.

Process - II is an isobaric process

(Pressure remain constant)

$W=P\Delta V=3P_0[3V_0-V_0]=6P_0{V}_0$

Process - III is an isochoric process

(Volume remain constant)

$\Delta Q=\Delta U+W$

W = 0

$\Delta Q=\Delta U$

Process - IV is an isothermal process

(Temperature remains constant)

$\Delta Q=\Delta U+W$

$\Delta U=0$

$\text{Hence C is the correct option}$