A barometer shows that an air pressure inside of a rocket before the launch is 10^5 pa. What is the change of air temperature inside the rocket during its launch, if the air pressure is increased to 1.2*10^5 pa ?

The rocket accelerates at a rate of 10 m/s^2.

###### Solution.

The process inside of the rocket is an isochoric process because a volume of the air inside is constant.

The equation of the isochoric process is

P1/T1=P2/T2 , where P1 is the initial pressure, P2 is the final pressure, T1 is the initial temperature and T2 is the final temperature.

Hence,

1×10^5 / T1 = 1.2×10^5 / T2

T2 / T1 = 1.2

Finally, the air temperature is increased 1.2 times.

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