Question

An ideal gas with 3 translational, 2 rotational and 2 vibrational degrees of freedom is at an initial temperature of 227 degrees Celsius, initial pressure of 24 atm and occupies an initial volume of 1800 L.

A) How many moles of molecules make up this gas? If the gas expands isothermally to a volume of 5400 L. What is the new pressure?

B) How much work was done in the process? What was the change in internal energy?

C) How much heat flowed into or out of the gas? The gas then undergoes isobaric contraction back to 1800 L. What is the new temperature?

D) How much work was done by the gas in the isobaric contraction (assume no phase change)? What was the change in internal energy of the gas?

E)How much heat flowed into or out of the gas in the contraction? The gas then pressurizes back to its initial pressure and temperature isochorically. What is the work done by the gas during the pressurization?

F) What was the change in internal energy during the pressurization? How much heat flowed into or out of the gas during the pressurization?

G) How much total work was done by the gas in this thermal cycle? How much heat flowed into the gas during this thermal cycle?

H) What was the efficiency of this thermal cycle? What would be the efficiency of a Carnot engine running in the same temperature range?

I) Sketch this three-process thermal cycle on a PV diagram. Indicate the pressure and volume at the endpoints of each process in your graph.

Please answer E,F,G,H,I.

Answer 1

Given that

T1=227+273=500K

P1=24 atm

v1=1800 litre

(A)to find the number of moles we calculate the volume of gas at STP condition

so at STP temperature and pressure are

T2=273 k and P2=1 atm

so by using ideal gas equation
P1V1/T1=P2V2/T2

=>24x1800/500=1xV2/273

=>V2=23587.2 L

we know that at STP volume of 1 mole of gas is 22.4 liters.So

22.4 L gas at STP=1 mole

23587.2L gas at STP=23587.2/22.4

                              =1053 moles

If the gas expands isothermally to the volume 5400 L then using Boyle's Law-

P1V1=P2V2

24x1800=P2x5400

P2(new pressure)=24x1800/5400=8 atm

(B)work done in isothermal process

        (where n=number of moles)

=1053x8.314x500xln(5400/1800)

=4808978.642 J

=4.81x10⁶ J

since internal energy is the function of temperature and in isothermal process temperature is constant so change in internal energy=0

(C)in isothermal process

so by first law of thermodynamics

Q=0+w

=4.81x10⁶ J

=4.81x10⁶ x0.24

=1.15x10⁶ calorie

since this is positive so heat is added to the system.

for isobaric process

V1/T1=V2/T2

5400/500=1800/T2

T2=1800x500/5400

=500/3=166.67K

(D)work done by gas in isobaric process

=P(V2-V1)

when the gas was isothermally expanded then its pressure increases to 8atm and again it isn expanded isobarically so now its pressure is still 8 atm..

8 atm=8x1.01x10⁵ pascal=8.08x10⁵ pascal

som work done

W=8.08x10⁵ x(1800-5400)x10^-3 J (10^-3 is multiplied to convert litre into cubic meter)

=-29088x10²

=-2.91x10⁶ J

negative sign shows that work is done on the gas not by the gas..

we know that

Cv=(f/2)R

here f=7

Cv=7R/2

now change in internal energy

=1053x(7R/2)x(166.67-500)

=-10213646.86 J

=-1.021x10⁷ J