A container with a volume of #8 L# contains a gas with a temperature of #180^o K#. If the temperature of the gas changes to #360 ^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #8 L# contains a gas with a temperature of #180^o   K#.  If the temperature of the gas changes to #360     ^o    K# without any change in pressure, what must the container's new volume be?

Nora Arzate · Accepted Answer

The new volume is 16L.

Let's identify our known and unknown variables.

The first volume we have is #"8L"#, the first temperature is #180K#, and the second temperature is #360K#. Our only unknown is the second volume.

We can obtain the answer using Charles' Law which shows that there is a direct relationship between volume and temperature as long as the pressure and number of moles do not change.

The equation we use is:

The volume should have units of liters and the temperature must have units of Kelvins. In our case, both have good units!

Now rearrange the equation and plug in the given values

#V_2 = (T_2 * V_1)/(T_1)#
#V_2 = (360cancel("K") * "8 L") / (180 cancel("K"))#
#V_2 = "16 L"#

When using the Kelvin scale, you do not put the degree symbol. You just write K.

Amelia Adams · Answer

undefined Using the combined gas law $ P_1V_1/T_1 = P_2V_2/T_2 $, where $ P $ is pressure, $ V $ is volume, and $ T $ is temperature:

Given:
$ V_1 = 8 $ L,
$ T_1 = 180 $ K,
$ T_2 = 360 $ K,
$ P_1 = P_2 $ (constant pressure).

We need to find $ V_2 $.

Substituting the given values into the combined gas law:

$ (8 \, \text{L}) \times (180 \, \text{K}) = V_2 \times (360 \, \text{K}) $

Solving for $ V_2 $:

$$ V_2 = \frac{(8 \, \text{L}) \times (180 \, \text{K})}{360 \, \text{K}} $$

$$ V_2 = \frac{1440 \, \text{L} \cdot \text{K}}{360 \, \text{K}} $$

$$ V_2 = 4 \, \text{L} $$

Therefore, the container's new volume must be $ 4 \, \text{L} $.