15.0 L of an ideal gas at 298 K and 3.36 atm are heated to 383 K with a new pressure of 6.00 atm. What is the new volume in liters?
since PV=kT, we know that PV/T is constant.
That means we want V such that
6V/383 = 3.36*15/298
V = 10.796
you should do a sanity check once you have an answer. The temperature has not changed by much, so wince the pressure has almost doubled, the volume will be less.
To solve this problem, we can use the ideal gas law formula:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's find the number of moles of the gas using the initial conditions:
P1 = 3.36 atm
V1 = 15.0 L
T1 = 298 K
Using PV = nRT, we can rearrange the equation to solve for n:
n = (P1*V1)/(R*T1)
We need to know the ideal gas constant, which is 0.08206 L*atm/mol*K.
Substituting the values into the equation:
n = (3.36 atm * 15.0 L) / (0.08206 L*atm/mol*K * 298 K)
n ≈ 0.682 mol
Now, let's find the new volume using the final conditions:
P2 = 6.00 atm
T2 = 383 K
Using PV = nRT, we can rearrange the equation to solve for V:
V2 = (n * R * T2) / P2
Plugging in the values:
V2 = (0.682 mol * 0.08206 L*atm/mol*K * 383 K) / 6.00 atm
V2 ≈ 17.54 L
Therefore, the new volume is approximately 17.54 liters.
Apologies for the incorrect calculation. You are correct, using the relationship PV/T = constant, we can solve for the new volume.
Initially, we have P1 = 3.36 atm, V1 = 15.0 L, and T1 = 298 K.
After heating, we have P2 = 6.00 atm and T2 = 383 K.
Using the relationship, we have:
P1V1/T1 = P2V2/T2
Plugging in the values:
(3.36 atm * 15.0 L) / 298 K = (6.00 atm * V2) / 383 K
Simplifying the equation:
(3.36 * 15.0) / 298 = (6.00 * V2) / 383
Solving for V2:
V2 = (3.36 * 15.0 * 383) / (298 * 6.00)
V2 ≈ 8.62 L
Therefore, the new volume is approximately 8.62 liters.