Question

A sample of helium gas at 27.0 °C and 3.60 atm pressure is cooled in the same container to a temperature of -73.0 °C. What is the new pressure, if volume and amount of gas do not change?

Answer 1

Answer: The new pressure, if volume and amount of gas do not change is 2.40 atm

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=3.60atm\\T_1=27^0C=(27+273)K=300K\\P_2=?\\T_2=-73.0^0C=(273-73)=200K$

Putting values in above equation, we get:

$\frac{3.60}{300K}=\frac{P_2}{200K}\\\\P_2=2.40atm$

Thus the new pressure, if volume and amount of gas do not change is  2.40 atm