Kinetic Energy of gas Molecules

In summary, the conversation discusses finding the total kinetic energy of a 1.0 mol sample of hydrogen gas with a temperature of 30 C and using two different equations to calculate it. The conversation also includes a question about how fast a 75 kg person would have to run to have the same kinetic energy. The final solution involves using the equation (5/2)kT and multiplying by Avogadro's number for the correct answer.
  • #1
Dulcis21
3
0
Hi, the question, I'm having problems with is this:

A 1.0 mol sample of hydrogen gas has a temperature of 30 C. What is the total kinetic energy of all the gas molecules in the sample? How fast would a 75 kg person have to run to have the same kinetic energy?
Relevant equations:
K = (3/2)kT
or K = (3/2)nRT

Ok, so I tried using both formulas:

(3/2)(1.38*10^-23)(30+273) = 6.2721 * 10^-21

Then, since its one mole and the question asks for the total KE, I multiplied this answer by Avogadro's number (6.02*10^23) and got

= 3775.8042 Joules.

For the second part, I would just set this number equal to 0.5mv^2 and solve for v but since I got the first one wrong, it doesn't quite work :(


I really need help; I thought I had the right idea but it's not working out. Thanks
 
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  • #2
Dulcis21 said:
A 1.0 mol sample of hydrogen gas has a temperature of 30 C. What is the total kinetic energy of all the gas molecules in the sample? How fast would a 75 kg person have to run to have the same kinetic energy?



Relevant equations:
K = (3/2)kT
or K = (3/2)nRT




Ok, so I tried using both formulas:

(3/2)(1.38*10^-23)(30+273) = 6.2721 * 10^-21

Then, since its one mole and the question asks for the total KE, I multiplied this answer by Avogadro's number (6.02*10^23) and got

= 3775.8042 Joules.
Your method is right. Using [itex]U = \frac{3}{2}nRT[/itex], U = 1.5*8.314*303 = 3778.7 Joules.

For the second part, I would just set this number equal to 0.5mv^2 and solve for v but since I got the first one wrong, it doesn't quite work
Think of one mole or 1 gram of hydrogen possessing 3778.7 J of kinetic energy. The effective average (rms) velocity is:

[tex]v = \sqrt{2E/m}[/tex]

AM
 
  • #3
Andrew Mason said:
Your method is right. Using [itex]U = \frac{3}{2}nRT[/itex], U = 1.5*8.314*303 = 3778.7 Joules.

Think of one mole or 1 gram of hydrogen possessing 3778.7 J of kinetic energy. The effective average (rms) velocity is:

[tex]v = \sqrt{2E/m}[/tex]

AM


I thought my method was right but my homework web-assign isn't accepting that answer :(
It asked for it in 2 sig figs so i put it in as 3.8 *10^3 but it still said no...
 
  • #4
Dulcis21 said:
I thought my method was right but my homework web-assign isn't accepting that answer :(
It asked for it in 2 sig figs so i put it in as 3.8 *10^3 but it still said no...

Oh I figured it out lol, it needed (5/2)kT to work and then multiply by Avogadro's number.
 
  • #5
Dulcis21 said:
Oh I figured it out lol, it needed (5/2)kT to work and then multiply by Avogadro's number.
Of course. H2 is diatomic.

AM
 

1. What is kinetic energy of gas molecules?

Kinetic energy of gas molecules is the energy that they possess due to their motion. It is a type of energy that is associated with the movement of particles.

2. How is kinetic energy related to temperature?

The kinetic energy of gas molecules is directly proportional to the temperature of the gas. As the temperature increases, the speed and motion of the molecules also increase, resulting in a higher kinetic energy.

3. What factors affect the kinetic energy of gas molecules?

The kinetic energy of gas molecules is affected by the mass and velocity of the molecules. Heavier molecules have a lower kinetic energy compared to lighter molecules, while faster moving molecules have a higher kinetic energy than slower moving molecules.

4. Can the kinetic energy of gas molecules be measured?

Yes, the kinetic energy of gas molecules can be measured using various instruments such as a thermometer or a calorimeter. It is usually measured in units of joules (J) or kilojoules (kJ).

5. How does the kinetic energy of gas molecules affect the behavior of gases?

The kinetic energy of gas molecules determines the average speed and motion of the molecules, which in turn affects the pressure, volume, and temperature of the gas. The higher the kinetic energy, the more energetic and active the gas molecules are, resulting in a higher pressure and temperature.

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