Question

a gas has a pressure of 749.9 mm hg at 12.5 °c. what is the temperature, in °c, at a pressure of 634.0 mmhg? round your answer to 1 decimal place.

Explanation:

Step1: Convert initial temperature to Kelvin

$T_1 = 12.5+ 273.15=285.65\ K$

Step2: Apply Gay - Lussac's Law ($\frac{P_1}{T_1}=\frac{P_2}{T_2}$)

$T_2=\frac{P_2T_1}{P_1}$
Substitute $P_1 = 749.9\ mmHg$, $T_1 = 285.65\ K$, and $P_2 = 634.0\ mmHg$ into the formula.
$T_2=\frac{634.0\times285.65}{749.9}$
$T_2=\frac{181002.1}{749.9}\approx241.4\ K$

Step3: Convert final temperature to Celsius

$t_2=T_2 - 273.15$
$t_2=241.4 - 273.15=-31.8^{\circ}C$

Answer:

$-31.8$