Question

A sample of gas in a closed container at a temperature of 100.0°C and 3.0 atm is heated to 300.0°C. What is the pressure of the gas at the higher temperature?

Answer 1

Answer:- 4.6 atm.

Solution:- In this problem, the volume is constant and the pressure is changing as the temperature is changed. It's based on Gay-Lussac's law which states that, "At constant volume, pressure of the gas is directly proportional to the kelvin temperature."

The equation used for solving the problems based on this law is:

`(P_1)/(T_1)=(P_2)/(T_2)`

Where,
`P_1` is initial pressure and
`P_2` is final pressure.

Similarly,
`T_1` is initial temperature and
`T_2` is final temperature.

`P_1` = 3.0 atm

`T_1` = 100.0 + 273 = 373 K

`T_2` = 300.0 + 273 = 573 K

`P_2` = ?

Let's plug in the values in the equation and solve it for final pressure:

`(3.0atm)/(373K)=(P_2)/(573K)`

`P_2=((3.0atm*573K))/(373K)`

`P_2=4.6atm`

So, the pressure of the gas at higher temperature is 4.6 atm.

Answer 2

Initial temperature of the gas = 100.0°C + 273 = 373 K

Initial pressure of the gas = 3.0 atm

Final temperature of the gas = 300.0°C + 273 = 573 K

According to Pressure Law,

P₁/T₁ = P₂/T₂

where P₁ and T₁ are the initial pressure and temperature respectively, and P₂ and T₂ are the final pressure and temperature respectively.

Plugging in the given data in Pressure Law we have,

3.0 atm/ 373 K = P₂/573 K

P₂ = (3.0 atm x 573 K)/ 373 K

P₂ = 4.6 atm

Thus, the pressure of the gas at the higher temperature is 4.6 atm.