A container with a volume of $25 L$ contains a gas with a temperature of $150^o K$ . If the temperature of the gas changes to $120 ^o K$ without any change in pressure, what must the container's new volume be?

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Answer 1 · 2016-12-18T15:05:45

The new gas volume would have decreased to 20 L.

Apply Charles's law which states that the volume of an ideal gas is directly proportional to its absolute (Kelvin) temperature.

In equation form: V = k T

Your question requires that we write this in the form of a proportion

$V_1/T_1 = V_2/T_2$

(Note that these ratios are equal, because both would equal the constant, k.)

Now, we are ready:

$V_1$ and $T_1$ are the initial conditions - 25 L and 150 K.

$T_2$ is the new (changed) temperature - 120 K

$25/150 = V_2/120$

Solving for $V_2$ , we get $V_2$ = $(25 xx 120) / 150$ = 20 L

A reduction in temperature has resulted in a reduction in volume.

A container with a volume of 25 L25 L contains a gas with a temperature of 150^o K150^o K. If the temperature of the gas changes to 120 ^o K120 ^o K without any change in pressure, what must the container's new volume be?