The gas inside of a container exerts $28 P a$ $28 Pa$ of pressure and is at a temperature of $270^{o} K$ $270 ^o K$ . If the pressure in the container changes to $60 P a$ $60 Pa$ with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2017-04-30T12:41:00

The new temperature is $= 578.6 K$ $=578.6K$

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 $P_{1} = 28 P a$ $P_1=28Pa$

The initial temperature is $T_{1} = 270 K$ $T_1=270K$

The final pressure is $P_{2} = 60 P a$ $P_2=60Pa$

The final temperature is

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

$= \frac{60}{28} \cdot 270 = 578.6 K$ $=60/28*270=578.6K$

The gas inside of a container exerts 28 Pa28Pa28 Pa of pressure and is at a temperature of 270 ^o K270oK270 ^o K. If the pressure in the container changes to 60 Pa60Pa60 Pa with no change in the container's volume, what is the new temperature of the gas?