Question

a gas has an initial volume of 24.6 l at a pressure of 1.90 atm and a temperature of 335 k. the pressure of the gas increases to 3.50 atm, and the volume of the gas increases to 31.3 l. what is the final temperature of the gas? o 143 k o 231 k o 485 k o 785 k

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: Rearrange the formula to solve for $T_2$

$T_2=\frac{P_2V_2T_1}{P_1V_1}$.

Step3: Substitute the given values

$P_1 = 1.90$ atm, $V_1=24.6$ L, $T_1 = 335$ K, $P_2 = 3.50$ atm, $V_2=31.3$ L.
$T_2=\frac{3.50\times31.3\times335}{1.90\times24.6}$.
First, calculate the numerator: $3.50\times31.3\times335 = 3.50\times10485.5=36699.25$.
Then, calculate the denominator: $1.90\times24.6 = 46.74$.
$T_2=\frac{36699.25}{46.74}\approx785$ K.

Answer:

785 K