A gas in a sealed container has a pressure of 100 kPa at a temperature of 27 C. If the pressure in the container is increased to 20 kPa, what is the new temperature?

32.4 C

Explanation: Gay-Lussac's Law or Third Gas Law states that for a constant volume, the pressure is directly proportional to absolute temperature: P alpha T; also stated as P/T = K, where K is a constant, and similarly, P1/T1 = P2/T2 .

Here P1= 100 kPa, P2=120 kPa (If pressure increases by 20 kPA), T1=27 C [Given]

Hence, T2= (T1*P2)/P1 = (27*120)/100=32.4 C [C is degree in Celcius]

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