A 301.0 mL sample of O2 is collected over water at 47.0 °C. The total pressure is 98.8 kPa. What is the volume of the O2 at STP in mL? (the vapor pressure of water at this temperature is 149 torr). Explain your procedure please!
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Hi Hassan!
Sam is right about all gases occupying a volume of 22.4L at standard temperature and pressure (0 degrees Celsius and 1 atm). However, looking at the other information you provided, I’m guessing you also want me to prove it. Your values look a little weird but here is a proof:
You are going to need to use the ideal gas equation PV=nRT where P represents pressure, V represents volume in liters, n represents moles, R represents the gas constant (depending on the units of pressure), and T represents the temperature in kelvin (K=C+273). We must use kelvins instead of celsius because all gas equations must be expressed in positve numbers (celsius has negative values while kelvins do not).
So let's first convert some of the given values so that we can plug them into the equation. 47 degrees celsius+273=320K, (301.0 mL)/(1000mL/1L)=0.301 L Now, we want only the pressure of the O2 so let’s convert our total pressure to torrs and subtract the pressure of the water vapor. 1 torr = 0.132022368 kpa and so (98.8 kPa)(1 torr/0.132022368 kpa)= about 748 torr. Now subtract the pressure of the water vapor (748-149=599 torr) and you will get the pressure of O2. Now lastly, convert the torrs to atms and then we can use it in our equation (599 torr/1 atm/760torr) = about 0.788 atms..
Now that you're done converting, plug all your known variables into the equation and solve for n or the mols of O2. (0.779atm)(0.301L)=n(0.08206 L*atm/mol*K)(320K). N=(0.788atm)(0.301L)/(0.08206 L*atm/mol*K)(320K). So n= about 9.03x10^-3 mols of O2
Now the last step is to find the volume of the O2 at STP. For any gas at STP, 1 mol of the gas will equal to 22.4 liters. This is a ratio that you must memorize. Use this to convert your newly found mols of O2 to your liters. (9.03x10^-3 mols O2)(22.4L O2/1 mol O2) = about 0.202 L of O2. Now knowing that 1000mL equals 1L, multiply our answer by 1000 and you will get 202mL. See how this value is pretty close to the STP value (224 mL)? That difference is as a result of the errors in the values.
I hoped this helped and feel free to ask more questions :)
Sam is right about all gases occupying a volume of 22.4L at standard temperature and pressure (0 degrees Celsius and 1 atm). However, looking at the other information you provided, I’m guessing you also want me to prove it. Your values look a little weird but here is a proof:
You are going to need to use the ideal gas equation PV=nRT where P represents pressure, V represents volume in liters, n represents moles, R represents the gas constant (depending on the units of pressure), and T represents the temperature in kelvin (K=C+273). We must use kelvins instead of celsius because all gas equations must be expressed in positve numbers (celsius has negative values while kelvins do not).
So let's first convert some of the given values so that we can plug them into the equation. 47 degrees celsius+273=320K, (301.0 mL)/(1000mL/1L)=0.301 L Now, we want only the pressure of the O2 so let’s convert our total pressure to torrs and subtract the pressure of the water vapor. 1 torr = 0.132022368 kpa and so (98.8 kPa)(1 torr/0.132022368 kpa)= about 748 torr. Now subtract the pressure of the water vapor (748-149=599 torr) and you will get the pressure of O2. Now lastly, convert the torrs to atms and then we can use it in our equation (599 torr/1 atm/760torr) = about 0.788 atms..
Now that you're done converting, plug all your known variables into the equation and solve for n or the mols of O2. (0.779atm)(0.301L)=n(0.08206 L*atm/mol*K)(320K). N=(0.788atm)(0.301L)/(0.08206 L*atm/mol*K)(320K). So n= about 9.03x10^-3 mols of O2
Now the last step is to find the volume of the O2 at STP. For any gas at STP, 1 mol of the gas will equal to 22.4 liters. This is a ratio that you must memorize. Use this to convert your newly found mols of O2 to your liters. (9.03x10^-3 mols O2)(22.4L O2/1 mol O2) = about 0.202 L of O2. Now knowing that 1000mL equals 1L, multiply our answer by 1000 and you will get 202mL. See how this value is pretty close to the STP value (224 mL)? That difference is as a result of the errors in the values.
I hoped this helped and feel free to ask more questions :)
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All gases occupy 22.4L at STP.