A container with a volume of #6 L# contains a gas with a temperature of #250^o K#. If the temperature of the gas changes to #320 ^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #6 L# contains a gas with a temperature of #250^o   K#.  If the temperature of the gas changes to #320     ^o    K# without any change in pressure, what must the container's new volume be?

Layla Ahmed · Accepted Answer

The new volume is #=7.68L#

Apply Charles' Law

#" Volume (L) "/" temperature (K) "= " constant"#, #"(at constant pressure")#

#V_1/T_1=V_2/T_2#

The initial volume is #V_1=6L#

The initial temperature is #T_1=250K#

The final temperature is #T_2=320K#

The final volume is

#V_2=T_2/T_1*V_1=320/250*6=7.68L#

Leo Abeyta · Answer

undefined To find the new volume of the container, we use the combined gas law equation:

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

Given:
- $ V_1 = 6 \, \text{L} $ (initial volume)
- $ T_1 = 250 \, \text{K} $ (initial temperature)
- $ T_2 = 320 \, \text{K} $ (final temperature)
- Pressure ($ P $) remains constant

Rearrange the equation to solve for $ V_2 $:

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

Since pressure remains constant, $ P_1 = P_2 $, so they cancel out.

$$ V_2 = \frac{V_1 \cdot T_2}{T_1} $$

Substitute the given values:

$$ V_2 = \frac{6 \, \text{L} \cdot 320 \, \text{K}}{250 \, \text{K}} $$

$$ V_2 = \frac{1920 \, \text{LK}}{250} $$

$$ V_2 = 7.68 \, \text{L} $$

So, the container's new volume must be 7.68 L.