# The gas inside of a container exerts 18 Pa of pressure and is at a temperature of 60 ^o K. If the pressure in the container changes to 42 Pa with no change in the container's volume, what is the new temperature of the gas?

Jun 10, 2016

$140 K$.

#### Explanation:

From the general gas Laws, Boyle's Law, etc we get that :

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

$\therefore \frac{18}{60} = \frac{42}{T} _ 2$

$\therefore {T}_{2} = \frac{60 \times 42}{18} = 140 K$.