The gas inside of a container exerts $32 P a$ $32 Pa$ of pressure and is at a temperature of $50^{o} K$ $50 ^o K$ . If the temperature of the gas changes to $80^{o} K$ $80 ^oK$ with no change in the container's volume, what is the new pressure of the gas?

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Answer 1 · 2017-06-17T08:51:12

The new pressure is $= 51.2 P a$ $=51.2Pa$

We apply Gay Lussac's Law

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

The initial pressure is $P_{1} = 32 P a$ $P_1=32Pa$

The initial temperature is $T_{1} = 50 K$ $T_1=50K$

The final temperature is $T_{2} = 80 K$ $T_2=80K$

The final pressure is

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

$= \frac{80}{50} \cdot 32$ $=80/50*32$

$= 51.2 P a$ $=51.2Pa$

The gas inside of a container exerts 32 Pa32Pa32 Pa of pressure and is at a temperature of 50 ^o K50oK50 ^o K. If the temperature of the gas changes to 80 ^oK80oK80 ^oK with no change in the container's volume, what is the new pressure of the gas?