Some hydrogen gas is enclosed within a chamber being held at 200C with a volume of 0.025m^3. The chamber is fitted with a movable piston. Initially, the pressure in the gas is 1.50*10^6 Pa (14.8 atm). The piston is slowly extracted until the pressure in the gas falls to 0.95 * 10^6 Pa. What is the final volume V_2 of the container? Assume that no gas escapes and that the temperature remains at 200C.

How do you set this up? I know you use pV=nRT but the double pressure thing kinda throws me off

I got the above question correct, but this next one is the one I need help with:

Some hydrogen gas is enclosed within a chamber being held at 200C whose volume is 0.025m^3. Initially, the pressure in the gas is 1.50 * 10^6 Pa (14.8 atm). The chamber is removed from the heat source and allowed to cool until the pressure in the gas falls to 0.95 * 10^6 Pa. At what temperature T_2 does this occur?

since the temp is constant use p1v1=p2v2

no

To solve this problem, you need to use the ideal gas law equation, which is pV = nRT. This equation relates the pressure (p), volume (V), number of moles (n), gas constant (R), and temperature (T).

In this problem, the initial conditions are given as follows:
- Initial pressure (p1) = 1.50 * 10^6 Pa
- Initial volume (V1) = 0.025 m^3
- Temperature (T) = 200 °C (convert this to Kelvin)

To find the final volume (V2), we need to rearrange the ideal gas law equation as follows:

p1V1 = nRT
p2V2 = nRT

Since the number of moles (n) and temperature (T) remain constant throughout the process, we can equate the two equations:

p1V1 = p2V2

Now we can solve for V2:

V2 = (p1V1) / p2

Substituting the given values:
V2 = (1.50 * 10^6 Pa * 0.025 m^3) / (0.95 * 10^6 Pa)

Now, calculate the final volume V2 using a calculator:

V2 = 0.039 m^3

So, the final volume of the container is 0.039 m^3.