Collection of Complete Gas Kinetic Theory with Discussion
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Thus the pressure applied to the final state is 6.25 atm
c. 5 ρ
d. 6 ρ
e. 7 ρ
Discussion:
Is known:
P1 = P
T1 = T
ρ1 = ρ
P2 = 3P
T2 = 0.6T
Asked: ρ2 = ...?
Answer:
Based on the ideal gas equation obtained:
b. 8 kg
c. 12 kg
d. 20 kg
e. 25 kg
Discussion:
Is known:
m1 = 30 kg
P1 = 2 × 10⁵ Pa
T1 = 47 ° C = 47 + 273 = 320 K
P2 = 1.5 × 10⁵ Pa
b. 8 kg
c. 12 kg
d. 20 kg
e. 25 kg
Discussion:
Is known:
m1 = 30 kg
P1 = 2 × 10⁵ Pa
T1 = 47 ° C = 47 + 273 = 320 K
P2 = 1.5 × 10⁵ Pa
T2 = 27 ° C = 27 + 273 = 300 K
Asked: oxygen masses that escaped from the tube (Δm = ⋯?)
Answer:
The oxygen mass that escapes from the tube:
To find Δm, we must know the value of m2.
Based on the ideal gas equation obtained:
In the tube, the constant volume (mr and R is also constant) then the above equation can be written:
So:
So the oxygen mass that escapes from the tube:
Δm = m1- m2
Δm = 30 - 24 = 6 kg
b. 1440 K
c. 2400 K
d. 2800 K
e. 2650 K
Discussion:
Is known:
THe = 350 K
Mr. O2 = 32
Mr. He = 4
Asked: vrms He = vrms O2 if TO2 = ...?
Answer:
b. 48 seconds
c. 32 seconds
d. 9.6 seconds
e. 4.8 seconds
Discussion:
Is known:
V = 20 L = 2 × 10-² m³
P = 3.2 × 10⁵ Pa
P = 100 watts
Asked: time (t) for W machine = U monoatomic gas
Answer:
So the time required for W machine = U monoatomik gas
link : Collection of Complete Gas Kinetic Theory with Discussion
Collection of Complete Gas Kinetic Theory with Discussion - Here are some examples of questions and their discussion of the Gas Kinetic Theory.
1.A 40 L of oxygen at 47 ° C and 2.5 atm pressure is pressed so that the volume becomes 20 L and at the same time the temperature rises to 127 ° C, the applied pressure is ...
a. 4.25 atm
b. 6.54 atm
c. 8.70 atm
d. 6.25 atm
e. 12.65 atm
Discussion:
Is known:
V1 = 40 L = 0.04 m³
T1 = 47 ° C = 273 + 47 = 320 K
P1 = 2.5 atm
V2 = 20 L = 0.02 m³
T2 = 127 ° C = 273 + 127 = 400 K
Asked: P2 = ...?
Answer:
2. The mass of an ideal gas at temperature T and pressure P is ρ. If the gas pressure is tested 3P and the temperature is lowered to 0.6T, then the gas mass meeting in the latter state is ...
a. 0.6 ρ
b. 0.5 ρc. 5 ρ
d. 6 ρ
e. 7 ρ
Discussion:
Is known:
P1 = P
T1 = T
ρ1 = ρ
P2 = 3P
T2 = 0.6T
Asked: ρ2 = ...?
Answer:
Based on the ideal gas equation obtained:
So the gas mass density in the final state is ρ2 = 5 ρ1
3. A tank is filled with 30 kg of oxygen at a pressure of 2 × 105 Pa and a temperature of 47 ° C. The tank has a small hole so that a number of gases can pass out. On one day, when the temperature was 27 ° C and the pressure in the tank was 1.5 × 105 Pa, the oxygen mass that escaped out of the tank was ...
a. 6 kgb. 8 kg
c. 12 kg
d. 20 kg
e. 25 kg
Discussion:
Is known:
m1 = 30 kg
P1 = 2 × 10⁵ Pa
T1 = 47 ° C = 47 + 273 = 320 K
P2 = 1.5 × 10⁵ Pa
3. A tank is filled with 30 kg of oxygen at a pressure of 2 × 105 Pa and a temperature of 47 ° C. The tank has a small hole so that a number of gases can pass out. On one day, when the temperature was 27 ° C and the pressure in the tank was 1.5 × 105 Pa, the oxygen mass that escaped out of the tank was ...
a. 6 kgb. 8 kg
c. 12 kg
d. 20 kg
e. 25 kg
Discussion:
Is known:
m1 = 30 kg
P1 = 2 × 10⁵ Pa
T1 = 47 ° C = 47 + 273 = 320 K
P2 = 1.5 × 10⁵ Pa
T2 = 27 ° C = 27 + 273 = 300 K
Asked: oxygen masses that escaped from the tube (Δm = ⋯?)
Answer:
The oxygen mass that escapes from the tube:
To find Δm, we must know the value of m2.
Based on the ideal gas equation obtained:
In the tube, the constant volume (mr and R is also constant) then the above equation can be written:
So the oxygen mass that escapes from the tube:
Δm = m1- m2
Δm = 30 - 24 = 6 kg
4. The rate of rms of helium molecules at 350 K is equal to the rate of rms of oxygen molecules at the temperature ...
(The relative molecular mass of oxygen = 32, helium = 4)
a. 850 Kb. 1440 K
c. 2400 K
d. 2800 K
e. 2650 K
Discussion:
Is known:
THe = 350 K
Mr. O2 = 32
Mr. He = 4
Asked: vrms He = vrms O2 if TO2 = ...?
Answer:
5. Helium (He) a monoatomic gas, filling container with volume 20 L. The gas pressure is 3,2 × 10⁵ Pa. So the time it takes a 100 watt engine to produce the same amount of energy as the energy in this gas is ...
a. 96 secondsb. 48 seconds
c. 32 seconds
d. 9.6 seconds
e. 4.8 seconds
Discussion:
Is known:
V = 20 L = 2 × 10-² m³
P = 3.2 × 10⁵ Pa
P = 100 watts
Asked: time (t) for W machine = U monoatomic gas
Answer:
So the time required for W machine = U monoatomik gas
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