Question

a gas is cooled from 365 k to 285 k while its volume changes from 12.8 l to 9.9 l. the initial pressure of the gas is 1.9 atm. what is the final pressure of the gas, rounded to the nearest tenth? 1.3 atm 2.9 atm 1.9 atm 9.9 atm

Explanation:

Step1: Recall the combined - gas law formula

The combined - gas law is $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$, where $P_1$ is the initial pressure, $V_1$ is the initial volume, $T_1$ is the initial temperature, $P_2$ is the final pressure, $V_2$ is the final volume, and $T_2$ is the final temperature.

Step2: Identify the given values

$P_1 = 1.9$ atm, $V_1=12.8$ L, $T_1 = 365$ K, $V_2 = 9.9$ L, $T_2=285$ K.

Step3: Rearrange the formula to solve for $P_2$

$P_2=\frac{P_1V_1T_2}{T_1V_2}$.

Step4: Substitute the values into the formula

$P_2=\frac{1.9\times12.8\times285}{365\times9.9}$
$P_2=\frac{1.9\times12.8\times285}{3613.5}$
$P_2=\frac{1.9\times3648}{3613.5}$
$P_2=\frac{6931.2}{3613.5}\approx1.9$ atm

Answer:

1.9 atm