The gas inside of a container exerts $9 P a$ $9 Pa$ of pressure and is at a temperature of $35^{o} K$ $35 ^o K$ . If the temperature of the gas changes to $20^{o} K$ $20 ^oK$ with no change in the container's volume, what is the new pressure of the gas?

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Answer 1 · 2017-06-05T03:23:59

$5$ $5$ $Pa$ $"Pa"$

To solve this problem, we can use the pressure-temperature relationship of gases, illustrated by Gay-Lussac's law:

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $(P_1)/(T_1)=(P_2)/(T_2)$

Let's rearrange this equation to solve for the final pressure, $P_{2}$ $P_2$ :

$P_{2} = \frac{P_{1} T_{2}}{T_{1}}$ $P_2 = (P_1T_2)/(T_1)$

Plugging in known values, we have

$P_2 = ((9"Pa")(20cancel("K")))/(35cancel("K")) = color(red)(5)$ $color(red)("Pa"$

rounded to one significant figure, the lowest amount given in the problem.

The gas inside of a container exerts 9 Pa9Pa9 Pa of pressure and is at a temperature of 35 ^o K35oK35 ^o K. If the temperature of the gas changes to 20 ^oK20oK20 ^oK with no change in the container's volume, what is the new pressure of the gas?