How Does the Ideal Gas Law Apply to Changing Conditions in a Gas System?

[mustang]

Problem1. A cylinder with a movable piston contains gas at a temperature of 42 degrees Celicius, with a volume of 40m^3 and a pressure of 0.233*10^5 Pa.
What will be the final temperature of the gas if it is compressed to 0.728 m^3 and its pressure is increaded to 0.609*10^5 Pa? Answer in K.
How is this done?

Problem 3.
A gas bubble with a volume of 0.14 cm^3 is formed at the bottom of a 11.1 cm deep container of mercury. The temperature is 24 degrees Celisius at the bottom of the container and 43 degees Celisuis at the top of the container.
The acceleration of gravity is 9.81 m/s^2.
What is the volume of the bubble just beneath the surface of the mercury? Assume that the surface is at atmospheric pressure.
Answer in units of m^3.
How is correctly done?

 

[Doc Al]

Please do not double post. https://www.physicsforums.com/showthread.php?t=25941

 

[M98Ranger]

For Problem 1, we can use the ideal gas law, PV = nRT, to solve for the final temperature of the gas. We know that the initial temperature (T1) is 42 degrees Celsius, the initial volume (V1) is 40m^3, and the initial pressure (P1) is 0.233*10^5 Pa. We can also assume that the number of moles (n) remains constant since the gas is contained within a closed system.

Using the equation PV = nRT, we can rearrange it to solve for the final temperature (T2):

T2 = (P2V2)/(nR)

Where P2 is the final pressure (0.609*10^5 Pa) and V2 is the final volume (0.728 m^3).

Plugging in the values, we get:

T2 = (0.609*10^5 Pa * 0.728 m^3)/(n * R)

To find the value of n, we can use the ideal gas law again, but with the initial conditions:

P1V1 = nRT1

Plugging in the values, we get:

0.233*10^5 Pa * 40m^3 = n * R * (42+273.15) K

Solving for n, we get n = 4.01 moles.

Now we can plug in all the values into our original equation to find T2:

T2 = (0.609*10^5 Pa * 0.728 m^3)/(4.01 moles * 8.314 J/mol*K)

T2 = 111.6 K

Therefore, the final temperature of the gas will be 111.6 Kelvin.

For Problem 3, we can use the same equation PV = nRT to solve for the volume of the gas bubble just beneath the surface of the mercury. We know that atmospheric pressure (P1) is equal to 1 atm, and the temperature (T1) at the surface is 43 degrees Celsius, or 316.15 K. We can also assume that the number of moles (n) remains constant since the gas bubble is contained within a closed system.

Using the equation PV = nRT, we can rearrange it to solve for the final volume (V2):

V2 = (nRT2)/P2

Where T2 is