A 8.00 L cylinder containing SF6 at a pressure of 0.400 atm is connected by a valve to 91.0L cylinder containing 02 at 342 torr pressure. Both cylinders are at 8.00 C. Calculate the partial pressure (torr) of SF6 when the valve is opened and the temperature of the cylinders changes to 78.0 C.

Use PV = nRT to calculate mols SF6 in the 8.00 L cylinder and do the same with the 91.0 L cylinder to calculate mols oxygen. Then add to determine total mols of both gases, and use PV = nRT (remember to change T to 78.0 + 273) and determine pressure of the gases under the new volume (add the volumes, too). Don't forget to change 342 torr to atm. Post your work if you get stuck.

I slept on this problem over the night and realized early this morning that I didn't answer your question. There isn't anything wrong with what I wrote; I just didn't answer the question. You asked for the PARTIAL pressure of ONE of the gases; I gave you the TOTAL pressure of BOTH gases.

So you need not go through the above for both gases, just the one for which you want the partial pressure. Let me know if this isn't clear.

To solve this problem, we need to apply the ideal gas law and the concept of partial pressure.

The ideal gas law is given by:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

First, let's calculate the number of moles of SF6 in the 8.00 L cylinder.

Using the ideal gas law, we can rearrange the equation to solve for the number of moles:
n = PV / RT

Given:
P = 0.400 atm
V = 8.00 L
R = 0.0821 L·atm/(mol·K)
T = 8.00°C = 8.00 + 273.15 = 281.15 K

Plugging in the values, we get:
n(SF6) = (0.400 atm) * (8.00 L) / (0.0821 L·atm/(mol·K) * 281.15 K)

Calculating the expression, we find:
n(SF6) = 0.15 mol

Next, let's calculate the number of moles of O2 in the 91.0 L cylinder.

Using the same approach as before:
Given:
P = 342 torr
V = 91.0 L
R = 0.0821 L·atm/(mol·K)
T = 8.00°C = 8.00 + 273.15 = 281.15 K

Converting torr to atm: 342 torr * (1 atm / 760 torr) = 0.450 atm

Plugging in the values, we get:
n(O2) = (0.450 atm) * (91.0 L) / (0.0821 L·atm/(mol·K) * 281.15 K)

Calculating the expression, we find:
n(O2) = 4.07 mol

Now that we have the number of moles of SF6 and O2 in the cylinders, we can calculate the total pressure when the valve is opened.

The total pressure can be calculated using the mole fraction of each gas:

P(total) = P(SF6) + P(O2)

Where:
P(SF6) = partial pressure of SF6
P(O2) = partial pressure of O2

The mole fraction of SF6 can be calculated as:
X(SF6) = n(SF6) / (n(SF6) + n(O2))

And the mole fraction of O2 can be calculated as:
X(O2) = n(O2) / (n(SF6) + n(O2))

Now, using the mole fractions, we can calculate the partial pressure of SF6 when the valve is opened.

P(SF6) = X(SF6) * P(total)

Given:
T = 78.0°C = 78.0 + 273.15 = 351.15 K

We have already calculated n(SF6) and n(O2), so we can calculate the mole fractions:

X(SF6) = n(SF6) / (n(SF6) + n(O2))
X(SF6) = 0.15 mol / (0.15 mol + 4.07 mol)

Calculating the expression, we find:
X(SF6) ≈ 0.0363

Similarly, we can calculate the mole fraction of O2:
X(O2) = n(O2) / (n(SF6) + n(O2))
X(O2) = 4.07 mol / (0.15 mol + 4.07 mol)

Calculating the expression, we find:
X(O2) ≈ 0.964

Now, plugging in the values, we can find the partial pressure of SF6:

P(total) = P(SF6) + P(O2)
P(SF6) = X(SF6) * P(total)
P(total) = (P(SF6) / X(SF6))

Let's calculate P(total) first:

P(total) = (0.400 atm) + (0.450 atm)
P(total) = 0.850 atm

Now, let's calculate P(SF6):

P(SF6) = (0.0363) * (0.850 atm)
P(SF6) ≈ 0.031 atm

To convert this pressure to torr:
P(SF6) = 0.031 atm * (760 torr / 1 atm)

Calculating the expression, we find:
P(SF6) ≈ 23.6 torr

Therefore, the partial pressure of SF6 when the valve is opened and the temperature of the cylinders changes to 78.0°C is approximately 23.6 torr.