A rigid container is filled with chlorine gas. The gas has a pressure of 2.75 bar. The tank is then cooled down to -20.0oC at which point the pressure of the gas is now measured to be 1.48 bar. What was the original temperature (in oC) of the gas?

Original temperature (T1) = - 37.16°C

Explanation:

Given:

Gas pressure (P1) = 2.75 bar

Temperature (T2) = - 20°C

Gas pressure (P2) = 1.48 bar

Find:

Original temperature (T1)

Computation:

Using Gay-Lussac's Law

⇒ P1 / T1 = P2 / T2

⇒ 2.75 / T1 = 1.48 / (-20)

⇒ T1 = (2.75)(-20) / 1.48

⇒ T1 = -55 / 1.48

⇒ T1 = - 37.16°C

Original temperature (T1) = - 37.16°C

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