Question

In: Physics

1. you are given a .5 m^3 container containing 9 moles of an ideal diatomic gas...

1. you are given a .5 m^3 container containing 9 moles of an ideal diatomic gas at 0° C, and you want to allow it to reach room temperature (20° C) while staying at the same pressure. You have a choice of holding the volume constant while heating it (process A) and then slowly decompressing the gas isothermally (process B) or simply heating it while maintaining pressure (process C).

a) What is the pressure of the gas?

b) What is the change in internal energy of the gas in going from the initial state to the final state?

c) Calculate the work done and heat transferred during process A. note whether these transfer energy into or out of the gas.

d) Calculate the work done and heat transferred during process B. note whether these transfer energy into or out of the gas.

e) Calculate the work done and heat transferred during process C. note whether these transfer energy into or out of the gas.

f) How do the total work and heat transfer compare for the two methods for getting from the initial state to the final state (AB and C)?

Solutions

Expert Solution

PV = nRT

a) V1=0.5m3

n= 9 moles

T1 = 0 oC = 273 K

T2 = 20 oC = 293 K

a) P= nRT/V = 9*8.314*273/0.5 = 40854.99 Pa

b) Change in internal energy, Eint = nCvT

Cv = (5/2)R, for diatomic gas

Eint = 9*(5/2)*8.314*20 = 3741.3 J

c) Work done = 0 at constant volume

heat transferred = change in internal energy = 3741.3 J

d) At constant temperature, W = -nRTln(V2/V1) =-9*8.314*273*ln(0.54/0.5) =  -1572.12 J

Heat Transferred = -Work Done = 1572.12 J

e) At constant pressure,

Work done= -PV

P2 = P1 =P

V2/V1 = T2/T1

V2 = (293/273)*0.5 = 0.54 m3

V = 0.54-0.5 = 0.04 m3

Work done, WC = -40854.99*0.04 = -1634.2 J

Heat Transferred, QC = change in internal energy-work done = 3741.3+1634.2 = 5375.5 J

e) Total work done in process AB, WAB = -1572.12 J

Total heat transferred in process AB, QAB = 3741.3 + 1572.12 = 5313.42 J

So, WAB > WC

QAB < QC


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