What is the net work done by the gas in the overall process ?

Question

TestHub · Accepted Answer

Explain the question:

Find the net work done by the gas in a three-step process: isobaric heating, reversible adiabatic expansion, and isobaric cooling back to the initial volume.

Concept:

This question is based on adiabatic and isobaric process.

Solution:

(1) Step 1: Isobaric Heating (1 → 2)

Initial state (A): $T = 300 K$ , $P = 1 a t m$ . Let $V_{1} = V_{0}$ . From $P V = n RT$ : $V_{0} = \frac{1. R .300}{1} = 300 R$ (in appropriate units).

Heated to $T_{2} = 600 K$ at $P_{1} = 1 a t m$ . Volume doubles: $V_{2} = 2 V_{0}$ .

$W_{b y, 1} = P Δ V = n R Δ T = 1 \cdot R (600 - 300) = 300 R$ .

(2) Step 2: Adiabatic Expansion (2 → 3)

Starts at $V_{2} = 2 V_{0}$ , $T_{2} = 600 K$ to final volume $V_{3} = 42 V_{1} = 42 V_{0}$ .

Find final temperature $T_{3}$ using $T_{2} V_{2}^{γ - 1} = T_{3} V_{3}^{γ - 1}$ with $γ - 1 = 2/3$ :

$T_{3} = 600. (\frac{2 V _{0}}{4 2 V _{0}})^{\frac{2}{3}} = 600. (\frac{1}{2 ^{\frac{3}{2}}})^{\frac{2}{3}} = 600. \frac{1}{2} = 300 K$

$W_{b y, 2} = n C_{V, m} (T_{2} - T_{3}) = 1 \cdot 1.5 R (600 - 300) = 450 R$

(3) Step 3: Isobaric Cooling (3 → 4)

First find pressure $P_{3}$ at state 3: $\frac{P _{2} V _{2}}{T _{2}} = \frac{P _{3} V _{3}}{T _{3}} ⟹ \frac{1.2 V _{0}}{600} = \frac{P _{3} .4 2 V _{0}}{300} ⟹ P_{3} = \frac{1}{4 2} a t m$ .

Cooling at constant $P_{3}$ to volume $V_{4} = V_{0}$ .

Find $T_{4}$ : $\frac{V _{3}}{T _{3}} = \frac{V _{4}}{T _{4}} ⟹ \frac{4 2 V _{0}}{300} = \frac{V _{0}}{T _{4}} ⟹ T_{4} = \frac{300}{4 2} = \frac{75}{2} K$ .

$W_{b y, 3} = n R (T_{4} - T_{3}) = 1. R (\frac{75}{2} - 300) = \frac{75}{2} R - 300 R$ .

(4)

Total Net Work done by gas:

$W_{n e t, b y} = W_{b y, 1} + W_{b y, 2} + W (_{b y, 3})$

$300 R + 450 R + (\frac{75}{2} R - 300 R) = 450 R + \frac{75}{2} R$

Chemistry - Thermodynamics - I Question with Solution | TestHub

Options:

Doubts & Discussion