# The gas inside of a container exerts 15 Pa of pressure and is at a temperature of 60 ^o K. If the pressure in the container changes to 36 Pa with no change in the container's volume, what is the new temperature of the gas?

Oct 6, 2017

$144 K$

#### Explanation:

PV=nr∆T
V=(nr∆T)/P

${P}_{1} = 15 P a$
${T}_{1} = 60 K$
${P}_{2} = 36 P a$

${V}_{1} = {V}_{2}$
T_2=?

And, $n \text{ & } r$ will remain the same. So,

$\frac{\cancel{n r} 60 K}{15 P a} = \frac{\cancel{n r} {T}_{2}}{36 P a}$

$\implies {T}_{2} = \frac{36 \cancel{P a} \cdot 60 K}{15 \cancel{P a}}$

$\implies {T}_{2} = 144 K$