Translational Kenetic Energy

In summary: Now you know the moles in the balloon. Calculate the moles in the cylinder. ehildOh, I see! Thank you so much! I was able to solve it.In summary, the problem involves finding the total translational kinetic energy of helium atoms in a spherical balloon filled with helium under given conditions. The universal gas law is used to relate pressure, volume, temperature, and number of moles of the gas. The correct formula for average translational energy is 1/2 RT per degrees of freedom, or 3/2 RT. Expressing all quantities in basic units, such as converting kPa to Pa and cm to m, is necessary for accurate calculations. To solve part (ii), the
  • #1
roam
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Homework Statement



A spherical balloon is filled with helium atoms (Note 1 atm = 101.3 kPa).


(i) What is the total translational KE of the atoms if the balloon has a diameter 13.0 cm at 37.0 °C and the pressure inside the balloon is 121.6 kPa?

(ii) The above balloon was filled from a cylinder of volume 0.2 m3 containing helium gas of 190.0 atm and at the same temperature as in the balloon in (i). How many of the balloons in (i) can the cylinder inflate?

The Attempt at a Solution



(i) To find the translational kenetic energy I think I need to use the equation

[tex]\frac{2}{3} nRT[/tex]

But the problem is that the number of moles is not given to us. I tried finding it using the formula n=m/M, but that won't work because I know that the molar mass of helium is 4 g/mol but I don't know what the mass (m) is!

T=37.0 °C + 273.15 = 310.15 K

[tex]\frac{2}{3} n(8.314)(310.15)[/tex]

I can't go any further.
Any help is appreciated. :smile:
 
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  • #2
You know the pressure and temperature of the gas. The balloon is a sphere, the diameter is given. Can you determine the volume?

There is the universal gas law, which relates the pressure, volume, temperature and number of moles of the gas.

ehild
 
  • #3
ehild said:
You know the pressure and temperature of the gas. The balloon is a sphere, the diameter is given. Can you determine the volume?

There is the universal gas law, which relates the pressure, volume, temperature and number of moles of the gas.

ehild

Thank you... I get it. But there is still something wrong:

[tex]V=\frac{4}{3} \pi r^3[/tex]
[tex]V=\frac{4}{3} \pi (6.5)^3=1150.3[/tex]

[tex]PV=nRT[/tex]
[tex](121.6 kPa) \times 1150.3 = n (8.314) \times (310.15 K) [/tex]
[tex]n= 54.24[/tex]

[tex]\frac{2}{3}nRT= \frac{2}{3} (54.24) (8.314) (310.15)[/tex]
[tex]=92829.13 J[/tex]

But this is false because the correct answer must be 210.0 J. I tried different units for temprature and pressure etc but it didn't work... what's wrong?
 
  • #4
Use appropriate units: m^3, K, Pa.

Also you used a wrong formula for the average translational energy. It is 1/2 RT per degrees of freedom, so 3/2 RT.

ehild
 
  • #5
ehild said:
Use appropriate units: m^3, K, Pa.

Also you used a wrong formula for the average translational energy. It is 1/2 RT per degrees of freedom, so 3/2 RT.

ehild

Oops that was a typo.

But if I used Pa instead of KPa, the value for n will be huge:

n=54245.3

[tex]\frac{3}{2} nRT = \frac{3}{2} (54245.3)(8.314)(310.15)[/tex]

The answer will be a VERY huge number. It's very far from 210.0 J! :(
I used the right formula & the right units, why is my answer so wrong?
 
  • #6
Express all quantities in the basic units. kPa means 10^3 Pa, cm means 0.01 m. Use the units during your calculations.

For example,

[tex]
V=\frac{4}{3} \pi (0.065 m)^3=1.1503 \cdot 10^{-3} m^3
[/tex]

ehild
 
  • #7
ehild said:
Express all quantities in the basic units. kPa means 10^3 Pa, cm means 0.01 m. Use the units during your calculations.

For example,

[tex]
V=\frac{4}{3} \pi (0.065 m)^3=1.1503 \cdot 10^{-3} m^3
[/tex]

ehild

Thank you SO much! I got it.

Now, any hints on how to solve part (ii)? I absolutely have no clue how to approach this problem...
 
  • #8
Now you know the moles in the balloon. Calculate the moles in the cylinder.

ehild
 

What is translational kinetic energy?

Translational kinetic energy is the energy possessed by an object due to its motion in a straight line. It is dependent on the mass and velocity of the object.

How is translational kinetic energy different from other types of kinetic energy?

Translational kinetic energy is specific to the motion of an object in a straight line, while other types of kinetic energy, such as rotational kinetic energy, are specific to the motion of an object in a circular or rotational path.

How is translational kinetic energy calculated?

The formula for calculating translational kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is the velocity of the object in meters per second.

What are some real-life examples of translational kinetic energy?

Some examples of translational kinetic energy in everyday life include a moving car, a thrown ball, and a person running.

How is translational kinetic energy related to potential energy?

Translational kinetic energy and potential energy are two forms of mechanical energy, meaning they are related to the physical motion of an object. When an object is at rest, it has potential energy, and when it is in motion, it has translational kinetic energy. The two forms of energy can be converted back and forth, such as when a ball is thrown into the air, it has potential energy at its highest point and translational kinetic energy when it reaches the ground.

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