# You know that a gas in a sealed container has a pressure of 111 kPa at 231°C. What will the pressure be if the temperature rises to 475°C?

Nov 20, 2015

${P}_{2} = 165 k P a$

#### Explanation:

First, this is a sealed container which means that the volume is constant. Since the only change happening is the raise of temperature, therefore, the number of mole is also constant.

From the ideal gas law: $P V = n R T$ we can say that at $V = c t e$ and $n = c t e$ the pressure $P$ is directly proportional to the temperature $T$.

Therefore, $P = k T \mathmr{and} \frac{P}{T} = k$, where $k$ is a constant $k = \frac{n R}{V}$.

Thus, $\frac{{P}_{1}}{{T}_{1}} = \frac{{P}_{2}}{{T}_{2}} \implies {P}_{2} = \frac{{P}_{1} \times {T}_{2}}{{T}_{1}}$

${P}_{2} = \frac{111 k P a \times 748 \cancel{K}}{504 \cancel{K}} = 165 k P a$