Question

simple gas laws

1. a sample of gas has an initial volume of 4.20 l at a pressure of 755 mmhg. if the volume of the gas is increased to 7.10 l (at constant temperature), what is its pressure?

missed this? read section 6.3; watch kcv 6.3

Explanation:

Step1: Identify the Gas Law

This is a Boyle's Law problem, which states that for a fixed amount of gas at constant temperature, \( P_1V_1 = P_2V_2 \). Here, \( P_1 = 755 \, \text{mmHg} \), \( V_1 = 4.20 \, \text{L} \), and \( V_2 = 7.10 \, \text{L} \). We need to find \( P_2 \).

Step2: Rearrange the Formula

Rearrange Boyle's Law to solve for \( P_2 \): \( P_2=\frac{P_1V_1}{V_2} \).

Step3: Substitute the Values

Substitute \( P_1 = 755 \, \text{mmHg} \), \( V_1 = 4.20 \, \text{L} \), and \( V_2 = 7.10 \, \text{L} \) into the formula:

$$ P_2=\frac{755 \, \text{mmHg} \times 4.20 \, \text{L}}{7.10 \, \text{L}} $$

Step4: Calculate the Result

First, calculate the numerator: \( 755 \times 4.20 = 3171 \). Then divide by the denominator: \( \frac{3171}{7.10} \approx 446.62 \, \text{mmHg} \).

Answer:

The pressure of the gas is approximately \( 447 \, \text{mmHg} \) (or more precisely \( 446.6 \, \text{mmHg} \)).