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 Boyle's Law problem, \( P_1V_1 = P_2V_2 \) (constant temperature, \( T_1 = T_2 \)).
Given \( P_1 = 755 \, \text{mmHg} \), \( V_1 = 4.20 \, \text{L} \), \( V_2 = 7.10 \, \text{L} \), find \( P_2 \).

Step2: Rearrange the formula

From \( P_1V_1 = P_2V_2 \), solve for \( P_2 \): \( P_2 = \frac{P_1V_1}{V_2} \)

Step3: Substitute values

\( P_1 = 755 \, \text{mmHg} \), \( V_1 = 4.20 \, \text{L} \), \( V_2 = 7.10 \, \text{L} \)
\( P_2 = \frac{755 \times 4.20}{7.10} \)

Step4: Calculate the result

First, multiply \( 755 \times 4.20 = 3171 \)
Then, divide by \( 7.10 \): \( \frac{3171}{7.10} \approx 446.62 \, \text{mmHg} \) (rounded appropriately)

Answer:

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