# The gas inside of a container exerts 9 Pa of pressure and is at a temperature of 650 ^o K. If the temperature of the gas changes to 210 ^oC with no change in the container's volume, what is the new pressure of the gas?

Oct 22, 2016

9.46Pa

#### Explanation:

By ideal gas law volume remaining constant Pressure is proportional to Temperature in Kelvin scale

Given

$P 1 \to \text{Initial pressure of the gas} = 9 P a$

${T}_{1} \to \text{Initial Temperature of the gas} = 650 K$

P_2-> "Final pressure of the gas"=?

${T}_{2} \to \text{Final Temperature of the gas} = {210}^{\circ} C = 683 K$

By the ideal gas law

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

$\implies {P}_{2} = \frac{{P}_{1} {T}_{2}}{T} _ 1 = \frac{9 \times 683}{650} \approx 9.46 P a$