Question

A sample of oxygen gas initially at 0.97 atm is cooled from 21°C to -68°C at constant volume. What is its final pressure (in atm)?

Answer 1

Q144039

A sample of oxygen gas initially at 0.97 atm is cooled from 21°C to -68°C at constant volume. What is its final pressure (in atm)?

Solution:

In question we are told that the volume of the sample of oxygen gas is constant.

The oxygen gas initially is at pressure = 0.97 atm, and Temperature = 21 ⁰ C

After cooling the final temperature of oxygen gas is - 68⁰C.

We have to find the pressure of oxygen gas after cooling it.

So, P_I = 0.97 atm ; T_I = 21⁰ C = 21 + 273.15 = 294.15 K;

P_F = ? ; T_F = -68 ⁰C = -68 + 273.15 = 205.15 K;

The Gay-Lussac’s law  relates pressure and temperature at constant volume.

Gay-Lussac’s law  : At constant volume and for a given mass of gas, the pressure exerted by a gas is directly proportional to the Kelvin temperature.

$$P ∝ T ;$$

If we are given the intial and final state of the gas, then

$P_I /P_F = T_I / T_F$

Flipping both side of the equation we have.

$P_F / P_I = T_F / T_I$

multiply both the side by P_I , we have

$P_F / P_I * P_I = T_F /T_I * P_I$

$P_F = T_F / T_I * T_I$

plug the information given in the question, we have

P_F = 205.15 K / 294.15 K * 0.97 atm;

P_F = 0.697 * 0.97 atm;

P_F = 0.677 atm;

In question we are given all the quantities in 2 significant figure. So our final answer must also be in

2 significant figure.

Hence the final pressure of the oxygen gas is 0.68 atm ;