To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = moles
R = gas constant
T = temperature
First, let's solve for the initial pressure of the gas in the 2.0-L cylinder:
P1V1 = n1RT1
P1 = (n1RT1) / V1
P1 = (2.0 mol x 0.0821 L.atm/mol.K x 24.3 K) / 2.0 L
P1 = 4.0 atm
Now, let's calculate the total moles of gas after 0.5 moles escape into the 3.0-L vacuum chamber:
n_total = n1 + n2
n_total = 2.0 mol - 0.5 mol
n_total = 1.5 mol
Next, let's calculate the pressure of the gas in the 3.0-L vacuum chamber using the final total moles and new volume and temperature:
P2 = (n_total x R x T2) / V2
P2 = (1.5 mol x 0.0821 L.atm/mol.K x 298 K) / 3.0 L
P2 = 4.09 atm
Therefore, the final pressure of the gas in the second chamber is 4.09 atm.
The valve of a 2.0-L cylinder containing 2.0 moles of a gas at 2.0 am and 24.3 K is opened, allowing 0.5 moles to escape into a 3.0-L vacuum chambe at 298 K. What is the final pressure of the gas in the second chamber?
• 0.245 atm
O 4.09 atm
• 16.3 atm
• 24.5 atm
1 answer