A container with a volume of #42 L# contains a gas with a temperature of #150^o K#. If the temperature of the gas changes to #75 ^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #42 L# contains a gas with a temperature of #150^o   K#.  If the temperature of the gas changes to #75     ^o    K# without any change in pressure, what must the container's new volume be?

Charles Aeschliman · Accepted Answer

#V_2 = 21# #"L"#

Given the temperature and volume of a gas, we are asked to calculate its final volume.

We can accomplish this by applying Charles' law, which illustrates the temperature-volume relationship of gases:

#ulbar(|stackrel(" ")(" "(T_1)/(V_1) = (T_2)/(V_2)" ")|)#

where

#T_1# and #T_2# are the initial and final absolute temperatures of the gas system (which must be in units of Kelvin)

#V_1# and #V_2# are the initial and final volumes of the gas

We are aware of:

#V_1 = 42# #"L"#

#T_1 = 150# #"K"#

#V_2 = ?#

#T_2 = 75# #"K"#

Let's rearrange the equation to solve for the final volume*, #V_2#:

#V_2 = (T_2V_1)/(T_1)#

Entering predetermined values:

#V_2 = ((75cancel("K"))(42color(white)(l)"L"))/(150cancel("K")) = color(red)(ulbar(|stackrel(" ")(" "21color(white)(l)"L"" ")|)#

The final volume of the gas container is thus #color(red)(21color(white)(l)"liters"#.

Emily Abbate · Answer

undefined To find the new volume of the container when the gas temperature changes from 150 K to 75 K without any change in pressure, you can use the combined gas law:

$$ \frac{{P_1 \times V_1}}{{T_1}} = \frac{{P_2 \times V_2}}{{T_2}} $$

Where:
- $ P_1 $ and $ P_2 $ are the initial and final pressures (which remain constant),
- $ V_1 $ is the initial volume of the container (42 L),
- $ V_2 $ is the final volume of the container (to be determined),
- $ T_1 $ and $ T_2 $ are the initial and final temperatures of the gas (150 K and 75 K, respectively).

Rearrange the equation to solve for $ V_2 $ and then plug in the given values to find the new volume of the container.