Question 1: Thermodynamics & KTG

Question

One mole of a monatomic 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 6P0V0
![](https://imagex.cdn.examgoal.net/agxEKBwKUYWG9lMV4/gNcA8RFoJUQu37FlNeHSLMCgNIEZH/R0HriGSprj4Eb3WDw7NOOp/image.png)

Accepted Answer

Process I is adiabatic, therefore, $\Delta$Q = 0
$\Rightarrow$ $\Delta$Q = $\Delta$U + W $\Rightarrow$ W = $-$$\Delta$U
Thus, W  0 Therefore, temperature of gas increases and no heat is exchanged between gas and surroundings.
Thus, the correct mapping is P $	o$ 3.
Process II is isobaric, therefore, pressure remains constant
$\Rightarrow$ W = P$\Delta$V = 3P
0
[3V
0
$-$ V
0
] = 6P
0
V
0
Therefore, work done by gas in process II is 6P
0
V
0
.
Thus, the correct mapping is Q $	o$ 4.
Process III is isochoric, therefore, volume remains constant
$\Rightarrow$ $\Delta$Q = $\Delta$U + W $\Rightarrow$ $\Delta$Q = $\Delta$U
and work done by gas is zero.
Thus, the correct mapping is R $	o$ 1.
Process IV is isothermal process, therefore temperature remains constant
$\Rightarrow$ $\Delta$Q = $\Delta$U + W $\Rightarrow$ $\Delta$Q = W and $\Delta$ = 0
Thus, the correct mapping is S $	o$ 2.
Therefore, option (C) is correct.

Answer

$P 	o 4;Q 	o 3;R 	o 1;S 	o 2$

Answer

$P 	o 1;Q 	o 3;R 	o 2;S 	o 4$

Answer

$P 	o 3;Q 	o 4;R 	o 1;S 	o 2$

Answer

$P 	o 3;Q 	o 4;R 	o 2;S 	o $

Thermodynamics & KTG

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