2 moles of a diatomic ideal gas undergothe following cyclic process, consistingof an isothermal, isobaric, and anadiabatic process. Given PA5 atm,Pg = Pc = 1 atm and VA = 2lt.%3D%3Da) Calculate the Volumes VB and Vc.b) Calculate the work WAB, WBc,WCAc) Calculate the Heat QAB, QBC, QCA•pMm)

Answered: 2 moles of a the followin of an isothe…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

2 moles of a diatomic ideal gas undergo
the following cyclic process, consisting
of an isothermal, isobaric, and an
adiabatic process. Given PA
5 atm,Pg = Pc = 1 atm and VA = 2lt.
%3D
%3D
a) Calculate the Volumes VB and Vc.
b) Calculate the work WAB, WBc,WCA
c) Calculate the Heat QAB, QBC, QCA•
pMm)

Expert Solution

Step 1

From the given conditions, the cyclic process would be graphically represented as below-

Step 2

Here,
AB is isothermal
BC is isobaric
CA is adiabatic

Also, for a diatomic gas, $γ = \frac{7}{5}$

(a)

Since AB is isothermal. So from ideal gas equation-

$\begin{array}{rcl} \Rightarrow & P_{A} V_{A} = P_{B} V_{B} \\ \Rightarrow & V_{B} = \frac{P_{A} V_{A}}{P_{B}} \\ \Rightarrow & V_{B} = \frac{5 \times 2}{1} \\ \Rightarrow & V_{B} = 10 ltr \end{array}$

Now, from the adiabatic process CA-

$\begin{array}{rcl} \Rightarrow & P_{A} {(V_{A})}^{γ} = P_{C} {(V_{C})}^{γ} \end{array}$
$\begin{array}{rcl} \Rightarrow & V_{C} = V_{A} {(\frac{P_{A}}{P_{C}})}^{\frac{1}{γ}} \end{array}$
$\begin{array}{rcl} \Rightarrow & V_{C} = 2 \times {(\frac{5}{1})}^{\frac{5}{7}} \end{array}$
$\begin{array}{rcl} \Rightarrow & V_{C} = 6.31 ltr \end{array}$

(b)

Work done in the isothermal process in going from A to B is-

$\begin{array}{rcl} W_{AB} & = & nRTln (\frac{V_{B}}{V_{A}}) \\ = & nR \times \frac{PV}{nR} \times \ln (\frac{V_{B}}{V_{A}}) \\ = & P_{A} V_{A} \times \ln (\frac{V_{B}}{V_{A}}) \\ = & 5 \times 2 \times \ln (\frac{10}{2}) \\ = & 16.09 J \end{array}$

Work done during the isobaric process in going from B to C is-

$\begin{array}{rcl} W_{BC} & = & P ∆ V \\ = & P_{B} [V_{C} - V_{B}] \\ = & 1 \times [6.31 - 10] \\ = & - 3.69 J \end{array}$

Work done during the adiabatic process in going from C to A is-

$\begin{array}{rcl} W_{CA} & = & \frac{1}{γ - 1} [P_{C} V_{C} - P_{A} V_{A}] \\ = & \frac{1}{\frac{7}{5} - 1} [1 \times 6.31 - 5 \times 2] \\ = & \frac{5}{2} \times (- 3.69) \\ = & - 9.225 J \end{array}$

(c)

In an isothermal process, change in internal energy is zero. Therefore-

$Q_{AB} = W_{AB} = 16.09 J$

In an isobaric process-

$\begin{array}{rcl} ∆ U_{BC} & = & {nC}_{V} ∆ T \\ = & n \times \frac{5}{2} R \times [T_{C} - T_{B}] \\ = & \frac{5 nR}{2} \times [\frac{P_{C} V_{C}}{nR} - \frac{P_{B} V_{B}}{nR}] \\ = & \frac{5}{2} \times [P_{C} V_{C} - P_{B} V_{B}] \\ = & \frac{5}{2} \times [(1 \times 6.31) - (1 \times 10)] \\ = & - 9.225 J \end{array}$

$\begin{array}{rcl} ∴ Q_{BC} & = & ∆ U_{BC} + W_{BC} \\ = & - 9.225 J + (- 3.69 J) \\ = & - 12.915 J \end{array}$

In an adiabatic process $∆ Q = 0$
$∴ Q_{CA} = 0$

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