By how many times does the volume of liquid oxygen evaporated into gaseous oxygen at a pressure of two kilograms increase?
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By how many times does the volume of liquid oxygen evaporated into gaseous oxygen at a pressure of two kilograms increase? (The temperature is at room temperature in the gaseous state)
When liquid oxygen in a unit volume is vaporized into oxygen under standard conditions, its volume becomes 799 times that of the original. Then, PV=nRT. At room temperature, the volume of a gas is only related to pressure and is inversely proportional to it; roughly speaking, at a pressure of 2 kilograms, it is 40 times that of liquid oxygen per unit volume. Rough estimate only; waiting to hear from fellow sea lovers
I checked and found that the density of liquid oxygen is 1.14 T/m3, while the density of gaseous oxygen under standard conditions is 1.419 KG/m3. The volume of gas obtained when it vaporizes under standard conditions increases by approximately 799 times. According to Pv=nRT, if the temperature difference is small and the pressure difference is 3 times, then the volume at a pressure of 2 kilograms should be 799/3 times the volume of liquid oxygen.
This can be calculated based on the densities of liquid and gaseous oxygen
PV=nRT is approximately 45 times, rounded off, haha
Yes, 800/3 is more or less acceptable
1 liter of liquid oxygen is equivalent to 804 Nm³ of oxygen gas, and the volume corresponding to 2 kilograms of it is then calculated
2 kg of gauge pressure, so 800/3=267 times.
Reply to 5# Chen Huandong: Many people have given their answers now, but none of them are the same. Could you give me your detailed calculation process?
Reply to 3# zengzhiyongyes: Many people have given their answers now, but none of them are the same. Could you show me your detailed calculation process?