5) Air, initially at 373 K , 1 atm , and V,=1 liter undergoes two successive reversible expansions -a reversible isothermal expansion to some intermediate volume V, , followed by a reversible adiabatic expansion to a final volume V, , cooling the gas to final temperature of 273 K. Calculate V,, V,,Q, and W, for the process. Assume ideal gas behavior.

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Take V2 final volume to be 3 liters.

5) Air, initially at 373 K , 1 atm , and V, = 1 liter undergoes two successive reversible expansions –a
reversible isothermal expansion to some intermediate volume V, , followed by a reversible adiabatic
expansion to a final volume V, , cooling the gas to final temperature of 273 K. Calculate V,, V,,Q, and
W, for the process. Assume ideal gas behavior.
Transcribed Image Text:5) Air, initially at 373 K , 1 atm , and V, = 1 liter undergoes two successive reversible expansions –a reversible isothermal expansion to some intermediate volume V, , followed by a reversible adiabatic expansion to a final volume V, , cooling the gas to final temperature of 273 K. Calculate V,, V,,Q, and W, for the process. Assume ideal gas behavior.
Expert Solution
Step 1

Given: T1=373K, P1=1atm, V1=1L, V2=3L, T2=273K

 

Gamma of air is 1.4.

As the first process is isothermal in nature, the temperature of the air remains constant. This, we have :

Tα=T1=373K

For the adiabatic process, we have : 

TVγ-1=constantT2V21.4-1=TαVα1.4-1273×30.4=373Vα0.4Vα0.4=1.136Vα=1.1362.5=1.375L

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