Question

a gas confined to a 462 cm³ container exerts a pressure of 107.4 kpa at 38.6 °c. what will the pressure be if the volume is expanded to 644 cm³ and the temperature remains 38.6 °c? which gas law equation should be used to solve this problem?

Explanation:

Step1: Identify the gas law

Since the temperature remains constant ($T_1 = T_2$), we use Boyle's law. Boyle's law is $P_1V_1=P_2V_2$.

Step2: Rearrange the formula to solve for $P_2$

From $P_1V_1 = P_2V_2$, we can solve for $P_2$ as $P_2=\frac{P_1V_1}{V_2}$.

Step3: Substitute the given values

Given $P_1 = 107.4$ kPa, $V_1=462$ cm³, $V_2 = 644$ cm³.
Substitute into the formula: $P_2=\frac{107.4\times462}{644}$.
First, calculate $107.4\times462 = 107.4\times(400 + 60+2)=107.4\times400+107.4\times60 + 107.4\times2=42960+6444+214.8 = 49618.8$.
Then, $P_2=\frac{49618.8}{644}=77.05$ kPa.

Answer:

The gas - law equation is $P_1V_1 = P_2V_2$. The pressure $P_2$ is $77.05$ kPa.