when a gas at 325 k and 767.0 mm hg changes in pressure to 810.0 mm hg, what is the new temperature? t = ? k

Explanation:

Step1: Identify the relevant gas - law

For a gas with constant volume, Gay - Lussac's law is used, which is $\frac{P_1}{T_1}=\frac{P_2}{T_2}$.

Step2: Assign values to variables

$P_1 = 767.0\ mmHg$, $T_1=325\ K$, $P_2 = 810.0\ mmHg$, and $T_2$ is the unknown.

Step3: Rearrange the formula to solve for $T_2$

$T_2=\frac{P_2T_1}{P_1}$.

Step4: Substitute the values into the formula

$T_2=\frac{810.0\ mmHg\times325\ K}{767.0\ mmHg}$. $T_2=\frac{263250}{767}\ K\approx343.2\ K$.

Answer:

$343.2$