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

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

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Solutions

Expert Solution

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/C_V

=1973/20.8=94.8 K

queen_honey_blossom answered 1 month ago

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