Ideal Gas Laws Problems and Solutions

 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:
T_A = 27⁰C = 300 K
N_A: N_B = 2 : 1
V_A: V_B = 3 : 1
Ideal gas equation
PV = NkTN_AT_A /V_A
 = N_BT_B/V_B
 (2)(300)/3 = (1)T_B/1
T_B = 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 227^o 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

We are faced with two conditions of air bubbles (gas). Therefore we set air bubbles at the bottom of the lake as state 2 and air bubbles on the surface of the lake as state 1. Thus, P₁ pressure is equal to the outside air pressure P₀. Pressure P₂ = P₀ + Ph, where Ph is the hydrostatic pressure by water at depth h.
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
P_h = 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