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?

May 24, 2016

$\text{The final temperature of gas is 18.21 L}$

Explanation:

${v}_{i} = 15 L \text{ Initial Volume of Gas}$
${T}_{i} = 280 \text{ "^o K" Initial Temperature of Gas}$

${T}_{f} = 340 \text{ "^o K" Final Temperature of Gas}$
V_f=? " Final Temperature of Gas"

${V}_{i} / {T}_{i} = {V}_{f} / {T}_{f}$

$\frac{15}{280} = {V}_{f} / 340$

${V}_{f} = \frac{15 \cdot 340}{280}$

${V}_{f} = 18.21 L$