Question #89308
Two identical gas systems, each containing 0.1 mole of ideal gas, are at 300 K and 3 atm pressure. One system is allowed to expand adiabatically and the other isothermally, till both attain normal pressure. Show that the volume of adiabatically expanded
system is less than the isothermally expanded system. The ratio of the heat capacities of the gas is
1.4.
1
Expert's answer
2019-05-17T11:19:26-0400

adiabatic equation

p1V1γ=p2V2γp_1V_1^{\gamma}=p_2V_2^{\gamma}


from this

(V1V2)1.4=1/3(\frac{V_1}{V_2})^{1.4}=1/3V2=311.4V1V_2=3^{\frac{1}{1.4}}V_1


isotherm equation

p1V1=p2V2p_1V_1=p_2V_{2'}

from this

V1V2=13\frac{V_1}{V_{2'}}=\frac{1}{3}V2=3V1V_{2'}=3V_1

In over

V2V2=30.41.4=1.37\frac{V_{2'}}{V_2}=3^{\frac{0.4}{1.4}}=1.37

V2=1.37V2V_{2'}=1.37\cdot V_2

So, in isotherm process volume 1.37 times more than volume adiabatic process


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