A `10.0 L` container is filled with a gas to a pressure of `2.00 atm` at `0^(@)C`. At what temperature will the pressure inside the container be `2.50 atm` ?

Explanation

As the volume of the container remains constant, applying pressure-temperature law,viz.
`(P_(1))/(T_(1))=(P_(2))/(T_(2))`
We get
`(2 atm)/(273 K)=(2.50 atm)/(T_(2))`
or `T_(2)=341 K=341-273^(@)C=68^(@)C`