D. Which flask has the higher total mass of gas molecules ? Explain. A sample of
ID: 704390 • Letter: D
Question
D. Which flask has the higher total mass of gas molecules ? Explain. A sample of Os gas initially at STP is compressed to a smaller volume at sonstant iempsraturs What effect does this change have on 13. A. the average kinetic energy of O; molecules? Explain B. the average speed of O molecules? Explain C. the total number of collisions with the container walls in a unit time? Explair 14. A. Calculate the rms speed, in m/s, for CO gas molecules at 25 °c. Answer: 515 m/s B. What is the ratio of this speed to that of Ar atoms at the same temperature ? Answer: ms speed CO-1.194 rms speed ArExplanation / Answer
13 A) The average kinetic energy of a gas molecule is given as
KEavg = ½*m*vavg2 = 3/2*k*T
where m is the mass of a O2 molecule; vavg is the average velocity of the gas molecules and T is the absolute temperature of the gas. k is the Boltzmann’s constant.
It is evident from the above expression that the average kinetic energy of gas molecules depends only on the absolute temperature of the gas. Since the gas is compressed to a lower volume at a constant temperature, hence, the average kinetic energy of the O2 molecules will remain unchanged.
B) It is evident from the above expression that the average kinetic energy of O2 molecules depends only on the temperature of the gas as
½*m*vavg2 = 3/2*k*T
The mass of the O2 molecule is constant. Since the temperature is held constant, the average kinetic energy and hence, average velocity of the O2 molecules remains unchanged.
C) The pressure of the O2 gas molecules is given as
P*V = n*R*T
where V is the volume of the O2 gas, n is the number of O2 molecules and T is the temperature of the O2 gas. Since the volume compression takes place at a constant temperature, hence, the pressure increases (the number of molecules n is kept constant).
The pressure of a gas is dependent on the number of collisions between the gas molecules and the walls of the container. Infact, the pressure of the gas is only due to the collisions between the molecules and the container. Since the pressure increases, hence, the number of collisions between the molecules and the container increase. Therefore, the number of collisions per unit time must increase.
14 A) The root mean square velocity of a gas is given as
vrms = ?3*R*T/M
where T is the absolute temperature of the gas and M is the molar mass of the gas. R is the gas constant expressed as 8.314 kg m2 s-2 K-1 mol-1.
Given the temperature of the gas is 25°C, we have T = (25 + 273) K = 298 K.
Also, the molar mass of CO is M = (1*12.011 + 1*15.999) g/mol = 28.01 g/mol = (28.01 g/mol)*(1 kg/1000 g) = 0.02801 kg/mol
Plug in values and get
vrms = ?3*(8.314 kg m2 s-2 K-1 mol-1)*(298 K)/(0.02801 kg/mol)
= ?(265359.3717 m2 s-2)
= 515.1304 m/s ? 515 m/s (ans).
B) The Graham’s law of diffusion can be used to find out the ratio of the speed of CO to the speed of Ar at a given temperature.
Therefore, we have.
(rms velocity of CO)/(rms velocity of Ar) = ?M2/M1
where M1 and M2 are the molar masses of CO and Ar respectively.
The molar mass of Ar is about 39.9 g/mol; therefore,
rms speed of CO/rms speed of Ar = ?(39.9 g/mol)/(28.01 g/mol)
====> rms speed of CO/rms speed of Ar = ?(1.42449)
====> rms speed of CO/rms speed of Ar = 1.1935 ? 1.194 (ans).
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.