Question

a 9.44 l sample of gas has a pressure of 0.787 atm and a temperature of 85°c. the sample is allowed to expand to a volume of 11.9 l and is cooled to 29°c. calculate the new pressure of the gas, assuming that no gas escaped during the experiment.

Explanation:

Step1: Convert temperatures to Kelvin

$$T_1 = 85 + 273 = 358\ K$$
$$T_2 = 29 + 273 = 302\ K$$

Step2: Use the combined gas law

The combined gas law is $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$. We need to solve for $P_2$.
Rearrange the formula to $P_2=\frac{P_1V_1T_2}{V_2T_1}$.
Substitute the values: $P_1 = 0.787\ atm$, $V_1 = 9.44\ L$, $T_1 = 358\ K$, $V_2 = 11.9\ L$, $T_2 = 302\ K$.
$$P_2=\frac{0.787\times9.44\times302}{11.9\times358}$$

Step3: Calculate the value

First, calculate the numerator: $0.787\times9.44 = 7.43928$, $7.43928\times302 = 2246.66256$.
Then, calculate the denominator: $11.9\times358 = 4260.2$.
$$P_2=\frac{2246.66256}{4260.2}\approx0.527\ atm$$

Answer:

$0.527\ atm$