The gas inside of a container exerts $6 P a$ $6 Pa$ of pressure and is at a temperature of $120^{o} K$ $120 ^o K$ . If the pressure in the container changes to $49 P a$ $49 Pa$ with no change in the container's volume, what is the new temperature of the gas?

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Answer 1 · 2018-03-21T14:42:52

The new temperature is $= 980 K$ $=980K$

Apply Gay Lussac's Law

$\frac{Pressure}{Temperature} = Constant$ $"Pressure"/"Temperature"= "Constant"$ , at $Constant Volume$ $"Constant Volume"$

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

The initial pressure is $P_{1} = 6 P a$ $P_1=6Pa$

The initial temperature is $T_{1} = 120 K$ $T_1=120K$

The final pressure is $P_{2} = 49 P a$ $P_2=49Pa$

The final temperature is

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

$= \frac{49}{6} \cdot 120 = 980 K$ $=49/6*120=980K$

The gas inside of a container exerts 6Pa6 Pa of pressure and is at a temperature of 120oK120 ^o K. If the pressure in the container changes to 49Pa49 Pa with no change in the container's volume, what is the new temperature of the gas?