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An ideal monatomic gas at an initial temperature of 500 K is expanded from 5.0 atm...

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Chemistry

An ideal monatomic gas at an initial temperature of 500 K isexpanded from 5.0 atm to a final pressure of 1.0 atm. Calculatew, q, DU, and (where applicable)DH and DT when the expansion is performed (a)reversibly and isothermally, and (b) reversibly andadiabatically.

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3.6 Ratings (464 Votes)

Given,

T=500 K

P1=5atm

P2=1 atm

a) For reversible isothermal process:

                w= -RT ln (P2/P1)

                =-8.314*500*ln(1/5)=6690.4 J

                q=-w=-6690.4 J

                dU=0

                dH=dU-w=0+6690.4=-6690.4 J

                dT= 0 for isothermal process

b) For reversible adiabatic process:

                w=-RT1((P2/P1­)(n-1)/n -1)

                =-8.314*500*((1/5)0.67/1.67-1)

                =-1973 J

                q=0 for adiabatic process

                dU=-w=1973 J

                dH=0

                dT= w/CV

                        =1973/20.8=94.8 K
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