The gas inside of a container exerts $15 P a$ $15 Pa$ of pressure and is at a temperature of $180^{o} K$ $180 ^o K$ . If the pressure in the container changes to $32 P a$ $32 Pa$ with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-04-17T11:32:02

The new temperature is $= 384 K$ $=384K$

We apply Gay Lussac's Law

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

Initial pressure is $P_{1} = 15 P a$ $P_1=15Pa$

Initial temperature is $T_{1} = 180 K$ $T_1=180K$

Final pressure is $P_{2} = 32 P a$ $P_2=32Pa$

Final temperature is

$T_{2} = \frac{P_{2}}{P_{1}} \cdot T_{1}$ $T_2=P_2/P_1*T_1$

$= \frac{32}{15} \cdot 180 = 384 K$ $=32/15*180=384K$

The gas inside of a container exerts 15Pa15 Pa of pressure and is at a temperature of 180oK180 ^o K. If the pressure in the container changes to 32Pa32 Pa with no change in the container's volume, what is the new temperature of the gas?