Question

a cylinder with a moveable piston contains a gas at 512 mm hg and 311 k at a constant volume. the pressure increases to 723 mm hg, what is the new temperature of the cylinder?

Explanation:

Step1: Identify the gas law

Since the volume is constant, we use Gay - Lussac's Law, which is $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, where $P_1$ is the initial pressure, $T_1$ is the initial temperature, $P_2$ is the final pressure, and $T_2$ is the final temperature.

Step2: List the given values

We have $P_1 = 512\space mm\space Hg$, $T_1=311\space K$, and $P_2 = 723\space mm\space Hg$. We need to solve for $T_2$.

Step3: Rearrange the formula to solve for $T_2$

From $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, we can cross - multiply to get $T_2=\frac{P_2\times T_1}{P_1}$.

Step4: Substitute the values into the formula

Substitute $P_1 = 512\space mm\space Hg$, $T_1 = 311\space K$, and $P_2=723\space mm\space Hg$ into the formula:
$T_2=\frac{723\space mm\space Hg\times311\space K}{512\space mm\space Hg}$
First, calculate $723\times311 = 723\times(300 + 11)=723\times300+723\times11 = 216900+7953=224853$
Then, divide by 512: $T_2=\frac{224853}{512}\approx439\space K$ (rounded to a reasonable number of significant figures)

Answer:

The new temperature of the cylinder is approximately $439\space K$