Question

1. choose the best answer.

a 500.00 ml sample of methane gas at 1.20 atm and 20.0 °c is cooled and compressed to 0.0 °c and 100.00 ml. what is its new pressure?
0.01 atm
5.59 atm
0.224 atm

Explanation:

Step1: Convert temps to Kelvin

$T_1 = 20.0^\circ\text{C} + 273.15 = 293.15\ \text{K}$
$T_2 = 0.0^\circ\text{C} + 273.15 = 273.15\ \text{K}$

Step2: Use combined gas law

The combined gas law is $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$. Rearrange to solve for $P_2$:
$P_2 = \frac{P_1V_1T_2}{V_2T_1}$

Step3: Substitute values

$P_2 = \frac{1.20\ \text{atm} \times 500.00\ \text{mL} \times 273.15\ \text{K}}{100.00\ \text{mL} \times 293.15\ \text{K}}$

Step4: Calculate final pressure

$P_2 = \frac{1.20 \times 500.00 \times 273.15}{100.00 \times 293.15} \approx 5.59\ \text{atm}$

Answer:

5.59 atm