Question

1. choose the best answer.

a sample of neon gas has a volume of 925 ml and a pressure of 1.57 atm at 25.0 °c. what is the volume if the temperature is lowered by 58 °c and the pressure is reduced to 0.815 atm?
1480 ml
3560 ml
2460 ml

Explanation:

Step1: Convert temps to Kelvin

Initial temp: $T_1 = 25.0 + 273.15 = 298.15\ \text{K}$
Final temp: $T_2 = (25.0 - 58) + 273.15 = 240.15\ \text{K}$

Step2: List known values

$V_1 = 925\ \text{mL}$, $P_1 = 1.57\ \text{atm}$, $P_2 = 0.815\ \text{atm}$

Step3: Apply combined gas law

Rearrange $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ to solve for $V_2$:
$V_2 = \frac{P_1V_1T_2}{P_2T_1}$

Step4: Substitute values

$V_2 = \frac{1.57\ \text{atm} \times 925\ \text{mL} \times 240.15\ \text{K}}{0.815\ \text{atm} \times 298.15\ \text{K}}$

Step5: Calculate final volume

$V_2 \approx 1480\ \text{mL}$

Answer:

1480 mL