Question

what pressure, in mm hg, will a sample of f₂ gas occupy at 225°c, if the gas exerts a pressure of 1260 mm hg at 323 k?
○ 1940 mm hg
○ 1810 mm hg
○ 817 mm hg
○ 878 mm hg

Explanation:

Step1: Identify the gas law

This is a gas law problem, specifically using Gay - Lussac's Law (since the volume of the gas sample is assumed to be constant, as we are dealing with pressure and temperature changes for a fixed amount of gas). Gay - Lussac's Law is given by the formula $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, where $P_1$ is the initial pressure, $T_1$ is the initial temperature, $P_2$ is the final pressure, and $T_2$ is the final temperature.

First, we need to convert the final temperature from Celsius to Kelvin. The formula for converting Celsius to Kelvin is $T(K)=T(^{\circ}C) + 273.15$. So, for $T_2 = 225^{\circ}C$, $T_2=225 + 273.15=498.15\ K$. The initial temperature $T_1 = 323\ K$ and the initial pressure $P_1 = 1260\ mm\ Hg$.

Step2: Rearrange the formula to solve for $P_2$

From $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, we can rearrange it to $P_2=\frac{P_1\times T_2}{T_1}$.

Step3: Substitute the values into the formula

Substitute $P_1 = 1260\ mm\ Hg$, $T_1 = 323\ K$, and $T_2 = 498.15\ K$ into the formula:

$P_2=\frac{1260\ mm\ Hg\times498.15\ K}{323\ K}$

First, calculate the numerator: $1260\times498.15 = 1260\times(500 - 1.85)=1260\times500-1260\times1.85 = 630000-2331 = 627669$

Then divide by the denominator: $P_2=\frac{627669}{323}\approx1943\ mm\ Hg$. Wait, but the closest option is 1940 mm Hg. Maybe we used a more precise or less precise conversion for temperature. Let's check with $T_2=225 + 273 = 498\ K$ (using 273 instead of 273.15 for simplicity).

Then $P_2=\frac{1260\times498}{323}=\frac{1260\times498}{323}=\frac{627480}{323}\approx1942.66\approx1940\ mm\ Hg$ (when rounded to a reasonable number of significant figures or to match the options).

Answer:

1940 mm Hg (the option with 1940 mm Hg)