Question

There is 0.5 L of air in a rigid, sealed container at a pressure of 203 kPa and a temperature of 203 K. The container is heated to 273 K.

What is the final pressure of the air?

Answer 1

Since, the container is rigid, there is no inflow as well as outflow of air.
Therefore, Pressure and Temperature remains constant.

Using Gay-Lussac's Law: The Pressure Temperature Law,
\[\frac{P1}{T1}=\frac{P2}{T2}\]
Plug the values into the equation,
\[\frac{203k}{203k}=\frac{P2}{273k}\]
\[P2=273k\]

Thus, final pressure of air would be 273kPa.

Answer 2

:D