Question

A person has a sample of gas with a volume of 7.41L and a temperature of 418.55K. If the volume of the container is reduced to 2.59L, what will the new temperature of the gas be in Kelvin?

Answer 1

For this problem, we could use the Charles's law:

`(V_1)/(T_1)=(V_2)/(T_2)`

We could replace the values that the question gives to us:

`(7.41L)/(418.55K)=(2.59L)/(T_2)\to T_2=(418.55K\cdot2.59L)/(7.41L)=146.29K`

Therefore, the new temperature of the gas is 146.29K.