what will be the pressure of the gas mixture when 0.5L of H2 at 0.8 bar and 2L of oxygen at 0.7 bar are introduced in a vessel at 27 degree C

We know the following:

• Volume of H2 = 0.5 L
• Pressure of H2 = 0.8 bar
• Volume of O2 = 2 L
• Pressure of O2 = 0.7 bar
• Temperature = 27°C (which is 300 K when converted to Kelvin)

Calculating Partial Pressures

First, we need to find the partial pressures of each gas in the same volume. To do this, we can use the ideal gas law, which is expressed as:

P = (nRT) / V

Where:

• P = pressure
• n = number of moles of gas
• R = ideal gas constant (0.0831 L·bar/(K·mol))
• T = temperature in Kelvin
• V = volume in liters

Finding Moles of Each Gas

We can rearrange the ideal gas law to find the number of moles (n) for each gas:

n = PV / RT

For Hydrogen (H2):

Using the values for hydrogen:

n(H2) = (0.8 bar * 0.5 L) / (0.0831 L·bar/(K·mol) * 300 K)

Calculating this gives:

n(H2) ≈ 0.0121 moles

For Oxygen (O2):

Now for oxygen:

n(O2) = (0.7 bar * 2 L) / (0.0831 L·bar/(K·mol) * 300 K)

Calculating this gives:

n(O2) ≈ 0.0564 moles

Calculating Total Moles

Now, we can find the total number of moles in the mixture:

n(total) = n(H2) + n(O2) ≈ 0.0121 + 0.0564 = 0.0685 moles

Finding Total Pressure of the Mixture

Next, we can find the total pressure of the gas mixture using the total number of moles and the ideal gas law again:

P(total) = (n(total) * R * T) / V(total)

Here, the total volume (V(total)) is the sum of the volumes of both gases:

V(total) = 0.5 L + 2 L = 2.5 L

Now substituting the values:

P(total) = (0.0685 moles * 0.0831 L·bar/(K·mol) * 300 K) / 2.5 L

Calculating this gives:

P(total) ≈ 0.68 bar

Final Result

Thus, the pressure of the gas mixture when 0.5 L of hydrogen at 0.8 bar and 2 L of oxygen at 0.7 bar are introduced into a vessel at 27°C is approximately 0.68 bar.