Question

A gas has a pressure of 2.36 kPa at 62 °C. What is the pressure at standard
temperature?

Answer 1

Answer:

P2 = 1.94 kPa

Explanation:

Given the following data;

Initial pressure = 2.36 kPa

Initial temperature = 62°C

Standard temperature = 0°C

Conversion:

Kelvin = 273 + C

Kelvin = 273 + 62 = 335 K

Kelvin = 273 + 0 = 273 K

To find the final pressure, we would use Gay Lussac's law;

Gay Lussac states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Gay Lussac's law is given by;

$PT = K$

$\frac{P1}{T1} = \frac{P2}{T2}$

Making P2 as the subject formula, we have;

$P_{2}= \frac{P1}{T1} * T_{2}$

$P_{2}= \frac{2.36}{335} * 273$

$P_{2}= 0.0071 * 273$

P2 = 1.94 kPa