# The gas inside of a container exerts 12 Pa of pressure and is at a temperature of 250 ^o C. If the temperature of the gas changes to 80 ^oK with no change in the container's volume, what is the new pressure of the gas?

Mar 31, 2016

$1.836 P a$ (4s.f.)

#### Explanation:

First convert temperature to Kelvin, $250 + 273 = 523 K$

Ideal gas law: ${P}_{1} {V}_{1} = n R {T}_{1}$
Calculate for number of moles:
$n = \frac{{P}_{1} {V}_{1}}{R {T}_{1}} = \frac{12 {V}_{1}}{8.314 \cdot 523} = 2.760 \times {10}^{-} 3 {V}_{1}$

Use Ideal gas law again:
${P}_{2} {V}_{2} = n R {T}_{2}$
Since ${V}_{2} = {V}_{1}$:
${P}_{2} {V}_{1} = 2.760 \times {10}^{-} 3 {V}_{1} \cdot 8.314 \cdot 80$
${V}_{1}$s cancel to leave:
${P}_{2} = 2.760 \times {10}^{-} 3 \cdot 8.314 \cdot 80 = 1.836 P a$