Question

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?

Answer 1

Answer: 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]

Answer 2

The new temperature of the gas in the container when the pressure increased by 20 KPa is 87 °C

Data obtained from the question

• Initial pressure (P₁) = 100 KPa
• Initial temperature (T₁) = 27 °C = 27 + 273 = 300 K
• New pressure (P₂) = 120 KPa
• New temperature (T₂) =?

How to determine the new temperature

The new temperature of the gas can be obtained as follow:

P₁ / T₁ = P₂ / T₂

100 / 300 = 120 / T₂

Cross multiply

100 × T₂ = 300 × 120

Divide both side by 100

T₂ = (300 × 120) / 100

T₂ = 360 K

Subtract 273 to obtain answer in °C

T₂ = 360 – 273 K

T₂ = 87 °C

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