Question #138534

A sample of gas at 25.00oC has a pressure of 92.50kPa. What temperature would change its pressure to 110.0kPa?

Expert's answer

We will use the relationship;

P1V1T1=P2V2T2\dfrac{P1V1}{T1}=\dfrac{P2V2}{T2}


From the question it seems Volume is kept constant. When that is the case, the relationship becomes;

P1T1=P2T2\dfrac{P1}{T1}=\dfrac{P2}{T2} (at constant volume)


P1=92.50kPa

P2 = 110.0kPa

T1 = 25 + 273 = 298K

T2 = ?


Solving for T2, the expression changes to;


T2=P2xT1P1T2 = \dfrac{P2 x T1}{P1}


T2=110.0kPax298K95.50kPa=343.246KT2 = \dfrac{110.0kPa x 298K}{95.50kPa} = 343.246K


Changing to oC, T2 = 70.25 oC

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