Question

A 1 L mixture of gases M and N is prepared by mixing gases from reservoirs of 10 L of volume each. The initial pressures on the reservoirs are 100 and 50 bar respectively. After mixing, the pressures in the reservoirs are 99 and 49 bar respectively. What is the pressure (in bar) of the gas mixture at 27°C?

• A. 2
• B. 1
• C. 20
• D. 10

Hint:

For gas mixtures, use the concept of moles transferred and scaling by volume to calculate partial pressures, then sum to get total pressure.

Answer 1

The Correct Option is C

Step 1: Recall the gas law.
For an ideal gas, \(P V = nRT\). The pressure is proportional to the number of moles when volume and temperature are constant.

Step 2: Determine the moles transferred.
Initial pressures: \(P_{M,i} = 100\) bar, \(P_{N,i} = 50\) bar.
Final reservoir pressures: \(P_{M,f} = 99\) bar, \(P_{N,f} = 49\) bar.
Hence, moles of M transferred: \(\Delta P_M = 100 - 99 = 1\) bar (equivalent).
Moles of N transferred: \(\Delta P_N = 50 - 49 = 1\) bar.

Step 3: Determine total moles in the mixture.
The 1 L mixture receives 1 bar equivalent from M and 1 bar equivalent from N. Total equivalent pressure contribution = 1 + 1 = 2 bar per liter
Reservoir volume = 10 L. Mixture volume = 1 L. Use partial pressure scaling:
Pressure contribution of M in 1 L mixture: \(P_M = \Delta P_M \cdot (V_{res}/V_{mix}) = 1 \cdot (10/1) = 10\) bar.
Pressure contribution of N in 1 L mixture: \(P_N = 1 \cdot (10/1) = 10\) bar.

Step 4: Total mixture pressure.
\[ P_{total} = P_M + P_N = 10 + 10 = 20 \text{ bar} \]

Step 5: Temperature check.
T = 27°C → constant

Step 6: Conclusion.
Hence, the pressure of the gas mixture at 27°C is \(\mathbf{20~bar}\).