Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.