A cook puts 7.20 g of water in a 2.00-L pressure cooker that is then warmed to 5
ID: 1557099 • Letter: A
Question
A cook puts 7.20 g of water in a 2.00-L pressure cooker that is then warmed to 550 degree C. What is the pressure inside the container? (Assume room temperature is 20 degree C.) 1.37 Assume that all the water has been converted to vapor. kPa. Calculate the mass of the air in a 2 m times 5 m times 2.5 m room, assuming that it is at 1 atm and 23 degree C. 3.690 Your responses differs from the correct answer by more than 10%. Double check your calculations. kg A 3.00-mol sample gas is confined to a 5.00-L vessel at a pressure of 7.96 atm. Find the average translational kinetic energy of the oxygen molecules under these conditions. 1/molecule A cylinder contains a mixture of helium and argon gas in equilibrium at 100 degree C. (a) What is the average kinetic energy for each type of gas molecule? Helium J argon J (b) what is the rms speed of each type of molecule? helium km/s argon m/sExplanation / Answer
problem: 5
(2.00 L) x (273 / (20 + 273)) / (22.414 L/mol) = 0.083139 mol air in the "empty" container
(7.20 g H2O) / (18.01532 g H2O/mol) = 0.39966 mol H2O added
Supposing all the water vaporized:
PV = nRT
P = nRT / V = (0.083139 mol + 0.39966 mol) x (0.08205746 Latm/Kmol) x (550 + 273 K) / (2.00 L) = 16.30 atm
16.30atm x 101.3kPa = 1651.19 kPa
problem 6:
The mass of a substance is equal to its density multiplied by its volume.
1L is defined to be exactly 1 dm^3 (one cubic decimeter), so
1L = 0.1m * 0.1m * 0.1m = 0.001 m^3
1.19 g/L = 1.19 / (0.001 m^3) = 1190 g/m^3
Remember that 1kg = 1000g, so
1190 g/m^3 = 1.19 kg/m^3
The volume of the room is the product of the width, length andheight:
V = 2.0 * 5.0 * 2.5 = 25.0 m^3
So:
mass = volume * density
= 25.0 m^3 * 1.19 kg / m^3
= 29.75 kg
problem 7:
First of all, we shall find the value of the temperature use the formula PV= nRT, where P is the pressure, V is the volume, n is the no of moles, R is the molar gas constant(8.31J/mol/K) and T is the absolute temperature.
1 atm= 1.01x10^5 Pa
7.96 atm= 7.96x1.01x10^5
= 8.0396x10^5 Pa
1L= 1x10^-3 m^3
5.01L= 5.01x1x10^-3
= 5.01x10^-3 m^3
T= PV/nR
= (8.0396x10^5)(5.01x10^-3)/(3.00)(8.314)
= 161.49K
Relative Molecular Mass(RMM) of O2= 32
1 mole of O2= 32g
2 moles of O2= 32x2
= 64g
We shall now use the formula 1/2(mv^2)= (3/2)kT, where m is the mass of O2 molecules, v is the root mean square velocity of O2 molecules k is the Boltzmann constant(1.38x10^-23) and T is the absolute temperature.
Rearranging it, we get v= (3kT/m)
= ((3)(1.38x10^-23)(161.49)/(64x10^-3))
= 3.23x10^-10 m/s
Average tran E= (1/2)mv^2
= (1/2)(64x10^-3)((3.23x10^-10)^2)
= 3.338x10^-21J
problem 8:
All (ideal) gas particles have the same mean KE at the same temperature (Kelvin)
From Kinetic Theory .. mean translational KE of particle = 3/2.kT .. (k=Boltzmann constant)
a) KE = 3/2 {1.38^-23 J/(K.particle)} (273+160)K
so, KE = 8.96x10^-21 J (both types)
b) KE = ½ m²
rms vel .. = (2 KE / m)
Taking the mass of a He atom .. m = 6.64x10^-27 kg
= (2 x 8.96x10^-21J / 6.64x10^-27kg)
or, (He) = 1642.80 m/s
Taking the mass of an Ar atom .. m = 66.40x10^-27 kg
= (2 x 8.96x10^-21 / 66.40x10^-27)
or, (Ar) = 519.50 m/s