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What is the pressure of 4 moles of helium in a 50L tank at 308K Use PVnRT?
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∙ 6y ago
2.02 atm
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What is the pressure of 4 moles of helium in a 50 liter tank at 308k?
2.02
How many moles of gas are present in a 10L container when 3.84atm of pressure is exerted at 35 Celsius?
Assuming that gas being used is an ideal gas, the gas law equation PV=nRT can be used. P=Pressure (atm) V=Volume (Liters) n=Moles R=0.08206 (a gas constant) T=Temperature (Kelvin) In this problem we know that there is a 10L container with 3.84atm of pressure at 35 C. So P=3.84atm, V=10L, and R=0.08206. To calculate T, you need to convert from Celsius (C) to Kelvin (K). Using the formula K=273+C, you should get T=308K. Now you know all of the variables you can work the problem out. Solve the equation for moles(n) by dividing each side by R and T: (PV/RT)=n Plug in your numbers and solve with a calculator: n=(3.84)(10)/(0.08206)(308) =(38.4)/(25.27448) =1.5193190918 Leaving us a rounded answer of 1.52n in the container
What is the pressure of 4 moles of helium in a 50 liter tank at 308k?
2.02
How many moles of gas are present in a 10L container when 3.84atm of pressure is exerted at 35 Celsius?
Assuming that gas being used is an ideal gas, the gas law equation PV=nRT can be used. P=Pressure (atm) V=Volume (Liters) n=Moles R=0.08206 (a gas constant) T=Temperature (Kelvin) In this problem we know that there is a 10L container with 3.84atm of pressure at 35 C. So P=3.84atm, V=10L, and R=0.08206. To calculate T, you need to convert from Celsius (C) to Kelvin (K). Using the formula K=273+C, you should get T=308K. Now you know all of the variables you can work the problem out. Solve the equation for moles(n) by dividing each side by R and T: (PV/RT)=n Plug in your numbers and solve with a calculator: n=(3.84)(10)/(0.08206)(308) =(38.4)/(25.27448) =1.5193190918 Leaving us a rounded answer of 1.52n in the container
10.0 grams of a gas occupies 12.5 liters at a pressure of 42.0 mm Hg What is the volume when the pressure has increased to 75.0 mm Hg?
Using Boyle's Law, p1*V1= p2*V2. This means that the pressure multiplied by the volume remains constant whilst the temperature is the same. Therefore; p1=42.0mm Hg, V1= 12.5L and so the product of the two is 525. If the pressure is now 75 mm Hg the volume must be 525/75= 7 liters. The 10.0 grams of gas information is not needed.
What is the volume in liters occupied by 1.21g of Freon-12 gas at 0.980 ATM and 308k?
V = m*R_sp*T/p = 1.21*(8.314/121)*308/(0.98*101325) = 2.58*10^-4 [m^3] = 0.258 [L]
