Question

One mole of an ideal monoatomic gas is taken around the cyclic process $A B C A$ as shown in figure. Calculate,

(a) the work done by the gas.

(b) the heat rejected by the gas in the path $C A$ and heat absorbed in the path $A B$ .

(c) the net heat absorbed by the gas in the path $B C$ .

Accepted Answer

(a) We are given a cyclic process of an ideal monatomic gas with the help of a diagram. The pressure and volume variations of the system are plotted which can be used to find the other parameters involved in this system during the process.

We know that the work done by the gas in the cyclic process can be easily found by finding the area under the pressure–volume curve of the process. Here, the area of the triangle $A B C$ will give us the work done $W$ as,

$C A$

$\Rightarrow W = \frac{1}{2} \times A B \times A C \Rightarrow W = 12 \times (3 P_{0} - P_{0}) \times (2 V_{0} - V_{0}) ∴ W = P_{0} V_{0}$

(b) Now, we can find the heat rejected by the process in $C A$ as,

$Q_{CA} = n C_{P} Δ T \Rightarrow Q_{CA} = n (\frac{5}{2} R) Δ T \Rightarrow Q_{CA} = n (\frac{5}{2} R) (P_{A} V_{A} - P_{C} V_{C}) ∵ P V = n R T Q_{CA} = n (\frac{5}{2} R) (P_{0} V_{0} - P_{0} 2 V_{0}) ∴ Q_{CA} = - \frac{5}{2} P_{0} V_{0}$

Also, the heat absorbed by the process along $A B$ is given similarly as,

$Q_{AB} = n C_{V} Δ T \Rightarrow Q_{AB} = n (\frac{3}{2} R) Δ T \Rightarrow Q_{AB} = n (\frac{3}{2} R) (P_{B} V_{B} - P_{A} V_{A}) ∵ P V = n R T \Rightarrow Q_{AB} = n (\frac{3}{2} R) (3 P_{0} V_{0} - P_{0} V_{0}) ∴ Q_{AB} = 3 P_{0} V_{0}$

(c) Now, the net heat absorbed by the gas along $B C$ is the work done by the gas along the path which is the area under the line $B C$ ,

$Q_{BC} = W_{BC} = area under B C \Rightarrow W_{BC} = (2 V_{0} - V_{0}) \times (3 P_{0} - P_{0}) ∴ W_{BC} = 2 P_{0} V_{0}$

Embibe Experts Solutions for Chapter: Thermodynamics, Exercise 3: BEGINNER'S BOX - 3

Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Thermodynamics, Exercise 3: BEGINNER'S BOX - 3

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