A gas cylinder holds 0.10 mol of O2 at 150°C and a pressure

of 3.0 atm. The gas expands adiabatically until the volume is doubled.

What are the final (a) pressure and (b) temperature?

Short Answer

Expert verified

The final (a) pressure is 1.1368 atm and (b) temperature is 321 K.

Step by step solution

01

Preparing the bases of the formulas

(a) This is an adiabic process. So, it can be wriiten that

p1V1γ=p2V2γp2p1=V1V2γp2=p1V1V2γ(1)

(b) The number of moles remains constant. From the ideal gas law,

p1V1T1=p2V2T2T2=p2p1V2V1T1

and from part (a),

p2p1=V1V2γ

Substitutiong this equation, it can be continue finding,

T2=V1V2γV2V1T1=V1V2γ-1T1(2)

02

Putting the values

(a)O2is a diatomic gas.

So, γ=1.40, Substituting this and the volume ratio, it can find from the equation (1) ,

p2=3·0.51.40=1.1368atm

(b) By putting the numerical values on the equation (2),

T2=0.51.40-1·423=321K

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