Answer:
variations of the same species
Explanation:
hope this helps have a good day
The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.
1.
Let's assume ideal gas.
Pressure for N₂: 2 bar*0.4 = 0.8 bar
Pressure for CO₂: 2 bar*0.5 = 1 bar
Pressure for CH₄: 2 bar*0.1 = 0.2 bar
2. For the volume, let's find the total volume first.
V = nRT/P = (1 mol)(8.314 J/mol-K)(30 +273 K)/(2 bar*10⁵ Pa/1 bar)
V = 0.0126 m³
Hence,
Volume for N₂: 0.0126 bar*0.4 = 0.00504 m³
Volume for CO₂: 0.0126*0.5 = 0.0063 m³
Volume for CH₄: 0.0126*0.1 = 0.00126 m³
Answer:
A. The rate of heat transfer through the material would increase.
Explanation:
To calculate the heat transfer in a heat exchanger you decide that there is not heat leakage to the surroundings, that means that magnitude of the two transfer rates will be equal. Any heat lost by the hot fluid, is gained by the cold fluid. The equation that describes this is Q = m×Cp×dT
Where:
heat = mass flow ×specific heat capacity × temperature difference
So if we increase the rate of flow of cooling water and the other variables that ypu can control remain the same, the result is that the rate of heat transfer through the material would increase, as it is stated in option a.
