Use the ideal gas law, expressed by the equation PV=nRT. The pressure, P, is 99.2 kPa. The temperature, T, is 45.0 degrees Celsius, which you must first convert to Kelvins by adding 273. The number of moles, n, is given as 0.55. The ideal gas constant, R, is approximately 8.3. Now, solve for volume, V:
99.2(V)=0.55(8.3)(45.0+273)
99.2(V)=1,451.67; V = 14.64 liters.
Use the combined gas law, PV = nRT
P = 10.634 atm
V = ?
n = 0.56 moles
R = gas constant = 0.0821 L-atm/deg-mole
T = 65ºC +273 = 338ºK
V = nRT/P = (0.56)(0.0821)(338)/10.634 atm
V = 1.46 liters = 1.5 liters (2 sig figs)
PV = nRT
P = Pressure (Pascals)
V = Volume
n = Number of Moles
R = Ideal Gas Constant
T = Temperature (K)
One mole takes 22.4l at STP.So 2.66mol takes a volume of 59.584l
The volume occupied by 2,50 moles of gas at 98,5 psi and 98,5 degrees F is 9,49 L.
The volume is 125,066 litres.
0,069 moles of gas
The volume is 6,19 L.
The volume is 13,64 L.
I would assume chlorine gas and standard temperature an atmospheric pressure. Using the ideal gas equation. PV = nRT (1 atm)(X volume) = (2.4 moles Cl2)(0.08206 Mol*K/L*atm)(298.15 K) Volume = 59 Liters of chlorine gas --------------------------------------------
First we need the moles of krypton gas. Monoatomic gas.0.405 grams Kr (1 mole Kr/83.80 grams)= 0.004833 moles krypton gas======================Now the ideal gas equation.PV = nRT(1 atm)(X volume) = (0.004833 moles Kr)(0.08206 L*atm/mol*K)(298.15 K)= 0.118 Liters of krypton gas--------------------------------------
340.89 k
The volume is 251,6 litres.
The volume is 3,81 L.
The volume is 6,19 L.
The volume is 13,64 L.
I would assume chlorine gas and standard temperature an atmospheric pressure. Using the ideal gas equation. PV = nRT (1 atm)(X volume) = (2.4 moles Cl2)(0.08206 Mol*K/L*atm)(298.15 K) Volume = 59 Liters of chlorine gas --------------------------------------------
what is the volume of a balloon containing 50.0 moles of O2 gas at a pressure of 15.0 atm at 28 degrees