Problem #1
A 590 liter volume tube containing oxygen gas at 20° C and a pressure of 5 atm. Determine the mass of oxygen in the tank! (Mr. Oxygen = 32 kg/kmol)Answer:
V = 5.9 x 10-1 m3
P = 5 x 1.01. 105 Pa
T = 20° C = 293 K
by using
PV = nRT and n = M/Mr so that:
PV = mRT/Mr
m = PVMr/RT
= 5 x 1.01 x 105 x 0.59 x 32/(8.314 x 293)
= 3.913 kg
Problem #2
The tank contains an ideal 3-liter gas with a pressure of 1.5 atm at 400 K. The gas pressure in the tank is raised at a fixed temperature up to 4.5 atm. Determine the volume of gas at the pressure!
Answer:
V1 = 3 liters; P1 = 1.5 atm; T1 = 400 K, P2 = 4.5 atm and T2 = 400 K
By using Boyle's law equation
P1V1 = P2V2
V2 = P1V1/P2
= 1.5 atm x 3 L/4.5 atm
= 1 liter
Problem #3
The air in a car tire at 15° C has a pressure of 305 kPa. After running at high speed, the tire becomes hot and the pressure becomes 360 kPa. What is the air temperature in the tire if the outside air pressure is 101 kPa?
Answer:
T1 = 288 K, P1 = 305 + 101 = 406 kPa, and P2 = 360 +101 = 461 kPa
By using the Gay-Lussac legal concept
P1/T1 = P2/T2
406/288 = 461/T2
T2 = 327 K = 54 ° C
Problem #4
16 grams of oxygen gas (M = 32 gr / mol) are at a pressure of 1 atm and a temperature of 270C. Determine gas volume? (given a value of R = 8.314 J / mol.K)
Answer:
R = 8,314 J/mol. K
T = 270C = 300 K
n = 16 gr: 32 gr/mol = 0.5 mol
P = 1 atm = 1.01 x 105 N/m2
By using the ideal gas equation
PV = nRT
(1,01 x 105 Pa)(V) = (0,5 mol)(8,314 Jmol-1K-1)(300 K)
(1,01 105Pa)(V) = 1249 J
V = 1249 J/(1.01 x 105 Pa) = 0.0125 m3
Problem #5
Tubes A and B are connected to a narrow pipe. The gas temperature at A is 270 C and the number of gas particles in A is twice the number of particles in B. If the volume is B one third of volume A, determine the temperature of the gas at B!
Answer:
TA = 270C = 300 K
NA: NB = 2 : 1
VA: VB = 3 : 1
Ideal gas equation
PV = NkTNATA /VA
= NBTB/VB
(2)(300)/3 = (1)TB/1
TB = 200 K
Problem #6
In a closed room a gas temperature is 27° C, a pressure of 1 atm and a volume of 0.3 liters. If the temperature of the gas is increased to 227° C and the pressure becomes 2 atm, then the volume of the gas becomes ....
Answer:
T1 = 27°C = 300 K
P1 = 1 atm
V1 = 0.3 liter
T2 = 227°C = 500 K
P2 = 2 atm
by using Boyle-Gay-Lussac law
P1V1/T1 = P2V2/T2
(1 atm)(0.3 L)/(300) = (2 atm)V2/(500 K)
V2 = 0.25 L
Problem #7
Volume V gas is in a closed space pressurized P and temperature T. If the gas expands isobarically so that the volume becomes 1/4 times the initial volume, then the ratio of the initial and final gas temperatures is ...
Answer:
P1 = P → 1
T1 = T → 1
Isobaris means the pressure is the same as
P1 = P2 → 1
The volume becomes 1/4 times the volume at first meaning:
V2 = 1 and V1 = 4
by using Boyle-Gay-Lussac law
P1V1/T1 = P2V2/T2
(1)(1)/(T1) = (1)(4)/(T2)
T1: T2 = 4 : 1
Problem #8
Gas with a mass of 7.2 kg with a temperature of 27° C is in a hollow tube. If the tube is heated to a temperature of 227o C, the tube expansion is ignored, specify:(a) the mass of gas remaining in the tube(b) the mass of gas coming out of the tube(c) mass ratio of gas coming out of the tube with the initial mass of gas.
Answer:
Initial gas mass of m1 = 7.2 kg
The remaining mass of m2 of gasThe mass of gas coming out of the tube
Δm = m1 - m2
(a) the mass of gas remaining in the tube
PV = nRTPV
= (m/Mr)RT
Because T is constant, and P is constant, then
m1T1 = m2T2
(4 kg)(300 K) = m2(500 K)
m2 = 4.32 kg
(b) the mass of gas coming out of the tube
Δm = m1 - m2Δm = 7.2 kg - 4.32 kg = 2.88 kg
(c) mass ratio of gas coming out of the tube with the initial mass of gas
Δm/m1 = 2.88/4.32 = 2/3
Problem #9
The volume of an air bubble at the bottom of a lake is 1.5 cm3. The depth of the lake is 102 m. What is the volume of the air bubble when it is just below the surface of the water? (outside air pressure = 75 cmHg; mercury density 13.6 g / cm3, density of water = 1 g / cm3).
Answer
The question sketch is shown in the picture above.
State 2: V2 = 1.5 cm3; h = 102 m
State 1: V1 =. . .?, P1 = P0 = 75 cmHg
Assume the temperature of the water is all the same T1 = T2.
Hydrostatic pressure is
Ph = 102 m water = 10200 cm water x (1 cmHg / 13.6 cm water) = 750 cmHg
P2 = P0 + Ph = 75 cmHg + 750 cmHg = 75 x 11 cmHg
Now we can calculate the volume of air bubbles when right on the surface of the water, V1 with
P1V1/T1 = P2V2/T2
(75)(V1)/(T) = (75 x 11)(1,5)/(T)
V1 = 16 cm3
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