The gas inside of a container exerts $12 P a$ $12 Pa$ 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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The gas inside of a container exerts $12 P a$ $12 Pa$ 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 · 2016-12-31T00:11:08

$P_{2} = 6.86 P a$ $P_2 = 6.86 Pa$

Explanation:

The ideal gas laws (Charles' Law) state:

$\frac{P_{1} \cdot V_{1}}{T_{1}} = \frac{P_{2} \cdot V_{2}}{T_{2}}$ $(P_1 * V_1)/T_1 = (P_2 * V_2)/T_2$

Where $P_{1},_{2}$ $P_1,_2$ are pressures – units don't matter in this case as long as they are consistent, because this is a ratio.
$V_{1},_{2}$ $V_1,_2$ are the corresponding volumes in Liters
$T_{1},_{2}$ $T_1,_2$ are the temperatures in degrees

With a constant container volume, Charles' Law becomes simply:
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $P_1/T_1 = P_2/T_2$

Rearrange for your known values:
$P_{2} = P_{1} \cdot \frac{T_{2}}{T_{1}}$ $P_2 = P_1 * T_2/T_1$

$P_{2} = 12 \cdot \frac{20}{35}$ $P_2 = 12 * 20/35$

$P_{2} = 6.86 P a$ $P_2 = 6.86 Pa$

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The gas inside of a container exerts $12 P a$ $12 Pa$ 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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Explanation:

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1 Answer

Explanation:

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The gas inside of a container exerts $12 P a$ $12 Pa$ 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?