A container with a volume of #15 L# contains a gas with a temperature of #280^o K#. If the temperature of the gas changes to #340 ^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #15 L# contains a gas with a temperature of #280^o   K#.  If the temperature of the gas changes to #340     ^o    K# without any change in pressure, what must the container's new volume be?

Leo Abeyta · Accepted Answer

#"The final temperature of gas is 18.21 L"# #v_i=15 L" Initial Volume of Gas"# #T_i=280 " "^o K" Initial Temperature of Gas"# #T_f=340" "^o K" Final Temperature of Gas"# #V_f=? " Final Temperature of Gas"# #V_i/T_i=V_f/T_f# #15/280=V_f/340# #V_f=(15*340)/280# #V_f=18.21 L#

Adrian Ayala · Answer

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

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

Given:
- Initial volume ($ V_1 $) = 15 L
- Initial temperature ($ T_1 $) = 280 K
- Final temperature ($ T_2 $) = 340 K
- Pressure remains constant

Let's solve for the final volume ($ V_2 $):

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

Since pressure remains constant, we can cancel it out:

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

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

Substituting the given values:

$$ V_2 = 15 \times \frac{340}{280} $$

$$ V_2 = 15 \times \frac{17}{14} $$

$$ V_2 = 15 \times 1.2142857 $$

$$ V_2 \approx 18.214 \, \text{L} $$

So, the container's new volume must be approximately $ 18.214 \, \text{L} $.