Question

a 8.2 l sample of gas has a pressure of 0.8 atm at a temperature of 259 k. if the temperature increases to 301 k, causing the volume to increase to 11.5 l, what is the new pressure? round your answer to the nearest tenth. atm

Explanation:

Step1: Recall the combined - gas law

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

$P_1 = 0.8$ atm, $V_1=8.2$ L, $T_1 = 259$ K, $V_2 = 11.5$ L, $T_2=301$ K.

Step3: Rearrange the combined - gas law to solve for $P_2$

$P_2=\frac{P_1V_1T_2}{V_2T_1}$.

Step4: Substitute the values into the formula

$P_2=\frac{0.8\times8.2\times301}{11.5\times259}$.
First, calculate the numerator: $0.8\times8.2\times301 = 0.8\times2468.2=1974.56$.
Then, calculate the denominator: $11.5\times259 = 2978.5$.
Now, $P_2=\frac{1974.56}{2978.5}\approx0.7$ atm.

Answer:

$0.7$ atm