How do I calculate the water vapor pressure? Webindex24.ch - News from the Web
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Hi, need help with the following task in a pressure cooker are 0.3l water. The container holds 6l. Which water vapor pressure is expected at 300 C? (Water vapor = ideal gas) My Result p = 13223.99 kPa (high or something?) The calculation method would not be bad, or at least how do I get to the mole. Thank you!
Assumption: 0.3 liter (at 20 C) water to be evaporated. The water vapor behaves like an ideal gas. The vapor occupies a volume of 6 liters at 300 C. To apply the equation of state of an ideal gas determine the number of moles of water. With the density of water rho = 0.99821 kg / l [1] gives the mass of water m = V * rho Molar mass of water, values from [3] M (H2O) = 2 M (H) + M ( 0) M (H2O) = 2 * 1.00794 g / mol + 15.9994 g / mol M (H2O) = 18.01528 g / mol moles of water n = m / M (H2O) n = V * rho / M (H2O) n = l * 0.3 998.21 g / L / (18.01528 g / mol), n = 16.62272 mol The temperature required in Kelvin T = (300 + 273.15) = KT 573 , 15 K The volume is in cubic meters requires V = 6 l V = 6 V dm = 0.006 m equation of state of an ideal gas [2] p V = n RT with the universal gas constant R = 8.314510 J / (mol K), the pressure for a given volume steamcalc and temperature p = n RT / V p = 16.62272 8.314510 mol * J / (mol K) * 573,15 K / (0.006 m ) p = 1.320249 * 10 ^ 7 Pa p = 132 , 0249 bar arises a pressure of 13.2 MPa equal to 132 bar one. Probably steamcalc möchter the teachers hear this answer, because he wants to practice the computation via the ideal gas law. More precisely executed looks: In the pot were initially 0.3 liters of water and 5.7 liters steamcalc of air. The air flow was neglected. The air as an ideal gas assumed yields a molar nl = 5.7 l / (22.4141 mol / l) nl = 0.2543042 mol The amount is small compared to the number of moles of water. The neglect was allowed. With the amount of air in a pressure of 13.4 MPa equal to 134 bar is derived. Water boils at 100 C and normal atmospheric steamcalc pressure (about 1 bar). Here a condition of 300 C and 143 bar is considered. A look at the vapor pressure curve [5] of water shows, the bar is approximately the vapor pressure at 300 C at 100. (After steamcalc the curve [5] under 100 bar, according to the Magnus formula in [4] at 110 bar.) So the water vapor can not reach bar at 300 C 134. The assumption that all the water has evaporated, was wrong. There is a balance be set: Part of the water evaporates until the vapor pressure is reached. The pressure can not be calculated using the ideal gas law, because this law ignores the fact that water can condense. The pressure steamcalc of the water vapor above the water in the pot is about 100 bar equal to 10 MPa. If the teacher has specified that the ideal gas law is to be used, then he does not expect this solution. steamcalc Leave a Reply
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Hi, need help with the following task in a pressure cooker are 0.3l water. The container holds 6l. Which water vapor pressure is expected at 300 C? (Water vapor = ideal gas) My Result p = 13223.99 kPa (high or something?) The calculation method would not be bad, or at least how do I get to the mole. Thank you!
Assumption: 0.3 liter (at 20 C) water to be evaporated. The water vapor behaves like an ideal gas. The vapor occupies a volume of 6 liters at 300 C. To apply the equation of state of an ideal gas determine the number of moles of water. With the density of water rho = 0.99821 kg / l [1] gives the mass of water m = V * rho Molar mass of water, values from [3] M (H2O) = 2 M (H) + M ( 0) M (H2O) = 2 * 1.00794 g / mol + 15.9994 g / mol M (H2O) = 18.01528 g / mol moles of water n = m / M (H2O) n = V * rho / M (H2O) n = l * 0.3 998.21 g / L / (18.01528 g / mol), n = 16.62272 mol The temperature required in Kelvin T = (300 + 273.15) = KT 573 , 15 K The volume is in cubic meters requires V = 6 l V = 6 V dm = 0.006 m equation of state of an ideal gas [2] p V = n RT with the universal gas constant R = 8.314510 J / (mol K), the pressure for a given volume steamcalc and temperature p = n RT / V p = 16.62272 8.314510 mol * J / (mol K) * 573,15 K / (0.006 m ) p = 1.320249 * 10 ^ 7 Pa p = 132 , 0249 bar arises a pressure of 13.2 MPa equal to 132 bar one. Probably steamcalc möchter the teachers hear this answer, because he wants to practice the computation via the ideal gas law. More precisely executed looks: In the pot were initially 0.3 liters of water and 5.7 liters steamcalc of air. The air flow was neglected. The air as an ideal gas assumed yields a molar nl = 5.7 l / (22.4141 mol / l) nl = 0.2543042 mol The amount is small compared to the number of moles of water. The neglect was allowed. With the amount of air in a pressure of 13.4 MPa equal to 134 bar is derived. Water boils at 100 C and normal atmospheric steamcalc pressure (about 1 bar). Here a condition of 300 C and 143 bar is considered. A look at the vapor pressure curve [5] of water shows, the bar is approximately the vapor pressure at 300 C at 100. (After steamcalc the curve [5] under 100 bar, according to the Magnus formula in [4] at 110 bar.) So the water vapor can not reach bar at 300 C 134. The assumption that all the water has evaporated, was wrong. There is a balance be set: Part of the water evaporates until the vapor pressure is reached. The pressure can not be calculated using the ideal gas law, because this law ignores the fact that water can condense. The pressure steamcalc of the water vapor above the water in the pot is about 100 bar equal to 10 MPa. If the teacher has specified that the ideal gas law is to be used, then he does not expect this solution. steamcalc Leave a Reply
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