Please provide the calculation process
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If the dry gas content in the converted gas is 50% (by volume) and the water vapor content is also 50% (by volume), with the pressure of the converted gas being 1.7 MPa (gauge pressure), then the partial pressure of water vapor (absolute pressure) is 0.9 MPa. Please calculate.
Under these conditions, the volume fraction of dry gas: volume fraction of water vapor = volume of dry gas: volume of water vapor = amount of substance of dry gas, n1: amount of substance of water vapor, n2 = 1:1. If they are ideal gases, then P1 = n1 * (RT/V) and P2 = n2 * (RT/V); thus, P1:P2 = n1:n2 = 1:1. P1 and P2 represent the partial pressures of the dry gas and water vapor respectively. Since P1 + P2 = 1.7 + 0.1 = 1.8, it follows that P1 = 0.9
Could the person upstairs please explain? P1 + P2 = 1.7 + 0.1 = 1.8, so P1 = 0.9. Why is 0.1 added? Is it some kind of backward calculation method? ? I don’t quite understand; could you please explain?
The requirement for the title is absolute pressure; I added 0.1 to the gauge pressure