1) An ideal gas occupies a volume of 0.60 m^3 at 5atm and 400K. What volume does it occupy at 4.0atm and a temp of 200K?
I know how to find the volume with the m^3 and the atm but it threw me off since it gives the 4.0 atm and the temp 200K.
The quantity P V/T remains constant when the number of moles remains the same.
Therefore
P1 V1/T1 = P2 V2/T2
"V2" is the final volume.
V2 = V1*(P1/P2)*(T2/T1)
= 0.60 m^3 * (5/4)*(200/400)
= ___ m^3
To solve this problem, you can use the ideal gas law, which states:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin
In this case, you are given the initial volume (V1), initial pressure (P1), initial temperature (T1), as well as the final pressure (P2) and final temperature (T2). You are asked to find the final volume (V2).
To solve for V2, you can rearrange the ideal gas law equation as follows:
V1/T1 = V2/T2 * (P2/P1)
To plug in the values:
V1 = 0.60 m^3
P1 = 5 atm
T1 = 400 K
P2 = 4.0 atm
T2 = 200 K
Now, let's substitute the values into the equation:
0.60 m^3 / 400 K = V2 / 200 K * (4.0 atm / 5 atm)
Next, simplify the equation:
0.0015 m^3/K = V2 / 200 K * (0.8)
To isolate V2, multiply both sides of the equation by 200 K:
0.3 m^3 = V2 * 0.8
Finally, divide both sides of the equation by 0.8 to solve for V2:
V2 = 0.375 m^3
Therefore, the gas occupies a volume of 0.375 m^3 at 4.0 atm and a temperature of 200 K.