Answered: A 2.30-mol ideal diatomic gas,…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

A 2.30-mol ideal diatomic gas, initially at temperature 123.0◦C, expands adiabatically to triple its volume. The same final state can be reached by an isothermal expansion followed by an isobaric cooling.

What is the work done by the gas during the adiabatic expansion? (A) 6.73 kJ; (B) 6.43 kJ; (C) 6.13 kJ; (D) 5.83 kJ; (E) 5.53 kJ.

What is the work done by the gas during the isothermal expansion? (A) 12.7 kJ; (B) 11.7 kJ; (C) 10.7 kJ; (D) 9.7 kJ; (E) 8.7 kJ.

What is the heat removed from the gas during the isobaric cooling? (A) 9.73 kJ; (B) 9.43 kJ; (C) 9.13 kJ; (D) 8.83 kJ; (E) 8.53 kJ.

Expert Solution

Step 1

Given,

No of moles, $n = 2.3 m o l$

The initial temperature, $T_{1} = 123^{0} C = 123 + 273.15 K = 396.15 K$

$V_{2} = V_{1}$

Step 2

For a diatomic gas,

$c_{p} = \frac{7}{2} R c_{v} = \frac{5}{2} R γ = \frac{c_{p}}{c_{v}} = \frac{\frac{7}{2} R}{\frac{5}{2} R} = \frac{7}{5}$

The gas constant,

$R = 8.314 J / m o l . K$

The relation between temperature and volume is,

$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} \frac{T_{2}}{396.15} = {(\frac{V_{1}}{3 V_{1}})}^{\frac{7}{5} - 1} T_{2} = 396.15 {(\frac{1}{3})}^{\frac{2}{5}} = 255.276 K$

Step 3

The work done by the gas during the adiabatic expansion,

$W = \frac{n R ∆ T}{γ - 1} = \frac{2.3 \times 8.314 \times (255.276 - 396.15)}{\frac{7}{5} - 1} = - 6734.55 J = - 6.734 k J (N e g a t i v e b e c a u s e w o r k d o n e b y t h e g a s)$

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