Two cylinders at 27C are connected by a closed stopcock ystem. One cyclinder contains 2.4 L of hydrogen gas at o.600 atm; the other cylinder contains 6.8L of helium at 1.40 atm. Assume valve take up no room.

what is the total pressure when the valve is open ?
(0.600 atm)(L)/(0.0821)(300K)=5.846x10^-2
n=(1.40 atm)(6.8L)/(0.0821)(300k)=3.865x10^-1
Total pressure =(n1+n2..)(RT/V)

And what is your question?

To find the total pressure when the valve is open, we first need to calculate the number of moles of hydrogen gas and helium gas in their respective cylinders.

Let's start with the hydrogen gas cylinder:
Using the ideal gas law equation (PV = nRT), we can calculate the number of moles (n) of hydrogen gas.
We are given:
Pressure = 0.600 atm
Volume = 2.4 L
Gas constant (R) = 0.0821 L·atm/mol·K
Temperature = 300 K

Using the rearranged equation n = (PV) / (RT), we can calculate the number of moles:
n(hydrogen) = (0.600 atm * 2.4 L) / (0.0821 L·atm/mol·K * 300 K) ≈ 0.0496 moles

Next, let's calculate the number of moles of helium gas in the other cylinder:
We are given:
Pressure = 1.40 atm
Volume = 6.8 L
Gas constant (R) = 0.0821 L·atm/mol·K
Temperature = 300 K

Using the same rearranged ideal gas law equation, we get:
n(helium) = (1.40 atm * 6.8 L) / (0.0821 L·atm/mol·K * 300 K) ≈ 0.379 moles

Now that we have the number of moles for each gas, we can find the total pressure when the valve is open by using the sum of their partial pressures.
Total pressure = (n(hydrogen) + n(helium)) * (RT / V)

Substituting the values, we get:
Total pressure = (0.0496 moles + 0.379 moles) * (0.0821 L·atm/mol·K * 300 K) / (2.4 L + 6.8 L)
Total pressure ≈ 0.428 moles * 0.0821 L·atm/mol·K * 300 K / 9.2 L
Total pressure ≈ 1.1848 L·atm / 9.2 L ≈ 0.129 moles

Therefore, the total pressure when the valve is open is approximately 0.129 atm.