Total points Name (please print) (last) (first) I. [25 pts] A sample of a monato

Question

Total points Name (please print) (last) (first) I. [25 pts] A sample of a monatomic ideal gas is confined to a sealed container PA with a movable piston. The questions below refer to the processes shown on the PV diagram at right. Processes I and 3 are isochoric (constant volume) changes from state W to state and state rto state z respectively. 2 and 4 are isobaric changes from state Mo state Yand state Z back to state respectively. Assume that E- 8.00 J. The pressure and volume of the gas in state z are P. and V (ie., Pr Po and V V's), The volume of the gas in state Wis 3V. The pressure of the gas in state ris P 3P. a H pts] Is the temperature of the gas in state X grearer than less than, or equal to the temperature of the gas in state ll? If itis equal. explain your reasoning. If it is not equal, determine the ratio of the temperatures and show your work. b. ptsl Determine the internal energy of the gas in state ron Joules) Explain your reasoning. 4 pts) Determine the internal energy of the gas in state Z (in Joules. Explain. d. [4 pts] The work done on the gas in Process 2 is +24.00 J. Determine the sign and absolute value of the work done on the gas during the cyclic process ll'or ll' Explain your reasoning

Explanation / Answer

a) state X

PV=nRT

T=PV/nR

state W

T'=P'V/nR

P'<P

T'<T

temperature at X is greater than temperature of Y

T/T'=P/P'=3P0/P0

Tx/Ty'=3:1

internal energy= nCv dT=nfR dT/2

f is degree of greedom, for monoatomic gas, f=3

consider n= 1

dT = chage in temperature

Let us consider from Z to W

AT Z

P0V0/nR=Tz

AT W

P03V0/nR=Tw

dT=2P0V0/nR

U= 2P0V0 J=18.00 (given in Question)

P0V0 =9J

b) internal energy= nCv dT=nfR dT/2

f is degree of greedom, for monoatomic gas, f=3

consider n= 1

dT = chage in temperature

Let us consider from W to X

AT W

P03V0/nR=Tw

AT X

3P03V0/nR=Tx

dT=6P0V0/nR

U=3 P0V0/2 J = 13.5 J

c)

internal energy= nCv dT=nfR dT/2

Let us consider from Y to Z

AT Y

3P0V0/nR=TY

AT Z

P0V0/nR=TZ

dT=P0V0/nR

U=2P0V0 J=18J

d) Work done= area of the cycle

=(3P0-P0)(3V0-V0)= 4P0V0

=36 J (positive)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.