Question #25290

A sample of nitrogen gas exerts a pressure of 9.80 atm at 32 °C. What would its temperature be (in °C) when its pressure is increased to 11.2 atm?

Expert's answer

Modified Clapeyron-Mendeleev law equation can be used for answering this question:

PV=nRT

P1V1=nRT1

P2V2=nRT2

P1/T1=P2/T2

T2=P2/(P1/T1)

So the answer is T2=11.2 atm*32 °C /9.80 atm = 36,6 C

PV=nRT

P1V1=nRT1

P2V2=nRT2

P1/T1=P2/T2

T2=P2/(P1/T1)

So the answer is T2=11.2 atm*32 °C /9.80 atm = 36,6 C

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