One mole of ideal monoatomic gas (v=5/3) is mixed with one mole of diatomic gas (γ=7/5). What is γ for the mixture? (γ denotes the ratio of specific heat at constant pressure, to that at constant volume.)
3/2
35/23
23/15
4/3
A
4/3
B
3/2
C
23/15
D
35/23
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Solution
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For a monoatomic gas:
Cv=3R2
For a diatomic gas:
Cv=5R2
Thereofore Cvavg=3R2+5R22=2R
Also Cp=Cv+R
⟹Cpavg=3R ⟹γ=CpavgCvavg=32
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