A container with a volume of #48 L# contains a gas with a temperature of #210^o K#. If the temperature of the gas changes to #80 ^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #48 L# contains a gas with a temperature of #210^o   K#.  If the temperature of the gas changes to #80     ^o    K# without any change in pressure, what must the container's new volume be?

Charles Aeschliman · Accepted Answer

Use Charles' Law to get a new volume of #18.3L#

Charles' Law for an ideal gas is that for a given mass and constant pressure, #V propT#.

To solve problems, this can be written as

#V_1/T_1 =V_2/T_2#

In this problem #V_1=48L#, #T_1=210K#, #T_2=80K# and #V_2=?#

#48/210=V_2/80#

Cross multiply (multiply the numerator of one side by the denominator of the other side)

#210*V_2=48*80#

#210V_2=3840#

Divide both sides by 210 #(210V_2)/210=3840/210#

#V_2=18.3L#

Layla Ahmed · Answer

undefined The new volume of the container must be 140 L.

Nicholas Agrusa · Answer

undefined Using the combined gas law, which states that the ratio of the initial volume to the initial temperature is equal to the ratio of the final volume to the final temperature, we can solve for the final volume.

Given:
Initial volume ($V_1$) = 48 L
Initial temperature ($T_1$) = 210 K
Final temperature ($T_2$) = 80 K

We need to find the final volume ($V_2$).

Using the combined gas law formula:

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

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

$$ V_2 = 48 \times \frac{80}{210} $$

$$ V_2 = 18.285 \, \text{L} $$

So, the container's new volume must be approximately $18.29$ liters.