Kinetic Theory of Gases problems and Solutions

 Problem #1

In a closed room there is a gas with a pressure of 3.2 x 10⁵ N/m². If the gas density is 6 kg/m³, what is the effective speed of each gas particle?
A. 400 m/s
B. 300 m/s
C. 250 m/s
D. 200 m/s
E. 150 m/s

Answer:
by using the equation
v_rms = (3P/ρ)^1/2
= [(3 x 3.2 x 10⁵)/(6)]^1/2v_rms = 400 m/s

Problem #2
Three moles of gas are in a space of 36 liters. Each gas molecule has a 5 x 10^-21 Joule kinetic energy.General gas constants = 8.315 J / mol. K and Boltzmann constant = 1.38 x 10^-23 J / K. Calculate gas pressure in the space!
A. 4.00 atm
B. 3.72 atm
C. 2.67 atm
D. 2.10 atm
E. 1.67 atm

Answer:
Is known Number of mol (n) = 3 moles Volume = 36 liters = 36 x 10^-3 m³ Boltzmann constant  (k) = 1.38 x 10^-23 J/K Kinetic energy (EK) = 5 x 10^-21 JouleR = 8.315 J/mol.Kby using the equation
EK = 3/2 kT
5 x 10^-21 = 3/2 x (1.38 x 10^-23)(T)
T = 241.5 K
Then the pressure P is determined using
PV = nRT
P(36 x 10^-3) = 3 x 8.315 x 241.5
P = 1,67 atm

Problem #3
Let P & Q are two sample of ideal gases of equal mole. Let T be the temperature of both the gas Let EP and E_Q are there total energy respectively. Let M_P and M_Q are these respective molecular mass. Which  of these is true
A. E_P > E_Q
B. E_P < E_Q
C. E_P = E_Q
D. E_P ≥ E_Q
E. E_P ≤ E_Q

Answer:
E_P = 3/2 nRT
E_Q = 3/2 nRT
E_P = E_Q

Problem #4
The velocities of the molecules are 2v, 3v, 4v, 5v & 6v. The rms speed will be
A. 15v
B. v(15)^1/2
C. v
D. 3,1v
E. 4,1v

Answer:
v_rms = (∑v²/N)^1/2
= [(4v² + 9v² + 16v² + 25v² + 36v²)/6]^1/2
v_rms = v(15)^1/2

Problem #5
An ideal gas A is there.Intial temperature is 27°C.The temperature of the gas is increased to 927°C.Find the ratio of final Vrms to the initial Vrms
A. 1 : 1
B. 2 : 1
C. 3 : 1
D. 1 : 4
E. 1 : 5

Answer:
v_rms = (3RT/M)^1/2
So it is proportional to Temperature
Now
T₁ = 27°C = 300 K
T₂ = 927°C = 1200K
So intial v_rms = k√300
Final v_rms = k√1200
ratio of final to intial = 2 : 1

Problem #6
Two absolute scales M and N have triple points of water defined as 200M and 350N. what is the relation between T_M and T_N.
A. T_M : T_N = 1 : 3
B. T_M : T_N = 2 : 3
C. T_M : T_N = 2 : 7
D. T_M : T_N = 4 : 7
E. T_M : T_N = 4 : 5

Jawab:
Given that on absolute scale
Triple point of water on scale M = 200M
Triple point of water on scale N = 350N
Also, triple point of water on Kelvin scale = 273.16 K
Now temperature on scale M and on scale N is equivalent to 273.16 K on absolute temperature scale.
Thus, value of one degree on absolute scale M = (273.16/200) K
Or,
Value of temperature T_M on absolute scale M = (273.16 x T_M)/200 
Similarly value of temperature T_N on absolute scale N = (273.16 x T_N)/350
Since T_M and T_N represent the same temperature
273.16 × T_M/200 = 273.16 × T_N/350
Or, T_M = 200T_N/350 = 4T_N/ 7

Problem #7
Which one is not the assumption in kinetic theory of gases
A. The molecules interact during the collison
B. The molecules of the gas are in continual random motion
C. The collison are of short duration
D. The molecules are tiny hard sphere undergoing inelastic collision
E. None of these

Jawab:
Answer is D

Problem #8
An ideal gas undergoes the process describe by equation
P = P₀ - bV²
Where P₀ , a are positive constant and V is the volume of one mole of gas
Maximum temperature attainable by gas
A. 3 P₀(P₀/2Rb)^1/2
B. (1/3R)(P₀/3b)^1/2
C. 2/3(P₀/R)(P₀/2)^1/2
D. 2/3 (P₀/R)(P₀/3b)^1/2
E. (1/3R)(P₀/b)^1/2

Answer:
P = P₀ - bV²
= P₀ – b(RT/P)²
PV = RT for one mole of gas
T = (1/Rb^1/2)P(P₀ – P)^1/2
= (1/Rb^1/2)(P₀P² – P³)
For T_max
dT/dP = 0
d(P₀P² - P³) /dP= 0
P = 2P₀/3
Putting this value
T_max = 2/3 (P₀/R)(P₀/3b)^1/2

Problem #9
Find the RMS speed of a sample of neon gas at 80.6° F.
A. 11.22 m/s
B. 12.27 m/s
C. 17.26 m/s
D. 19.26 m/s
E. 21.22 m/s

Answer:
First convert Fahrenheit to Celsius. °C = (°F- 32)/1.8
= (80.6 – 32)/1.8
= 48.6/1.8
= 27.0
Convert Celsius to Kelvin.
K = °C + 273
= 27.0 + 273
= 300 K
Substitute the known information into the equation for RMS speed and solv, we get,.
v_rms = (3RT/M)^1/2
= [3(8.31)(300)/(20)]^1/2
= 19.26 m/s
From the above observation we conclude that, the RMS speed of a sample of neon gas at 80,6° F would be 19.26 m/s.