Pf = 157.47 Kinetic Theory Homework Help: Raindrop Collisions and Ideal Gas Laws

Click For Summary
SUMMARY

The forum discussion revolves around solving kinetic theory problems related to raindrop collisions and ideal gas laws. Key calculations include determining the average force exerted by raindrops on a skylight, the pressure developed, and the root mean square (rms) speed of nitrogen and hydrogen molecules. The user correctly applied formulas such as F(ave) = -2mV/(2L/V) and P = F(average)/Area. The discussion confirms the user's understanding of the concepts and provides guidance on using the ideal gas law for further calculations.

PREREQUISITES
  • Understanding of kinetic theory principles
  • Familiarity with the ideal gas law (PV = nRT)
  • Knowledge of basic physics equations for force and pressure
  • Ability to perform unit conversions (e.g., grams to kilograms, cm to meters)
NEXT STEPS
  • Learn about the implications of inelastic collisions in fluid dynamics
  • Study the derivation and applications of the ideal gas law
  • Explore the concept of root mean square speed in kinetic theory
  • Investigate the relationship between temperature and molecular speed in gases
USEFUL FOR

Students studying physics, particularly those focusing on thermodynamics and kinetic theory, as well as educators seeking to enhance their understanding of gas laws and molecular behavior.

tido_29
Messages
1
Reaction score
0
Kinetic theory HW help :(

I am new here and found the site while trying to find a formula. I am having problems solving some questions. I worked out what i could but don't know if i did it right. Please Help :(

PROBLEM 1

A raindrop of mass (1mg) fall vertically at a constant speed of 10 m/s, striking a horizontal skylight at the rate of 1000 drops/s and draining off. The window is 15cm X 25cm. Assume the collisions are completely inelastic.

a) Calculate the magnitude of the average force of the raindrops on the window.

i used this equation. F(ave) = -2mV/(2L/V)

-2(1x10^-6kg)(10m/s) / ( 2 (.0375m^2)/(10m/s) ) = 2.66x10^-3 (is this right?)

b) what is the resuling pressure developed by the raindrop.

I know that P= F(average)/Area I know the area and F(average) is from above.



PROBLEM 2

The molar mass of N2 is 28g.

a) find the mass of 1 nitrogen molecule.

(1mol N2/ 28g N2) X ( 6.02x10^23 molecules/ mol) = 2.15x10^25 mol/g so each molecule weighs (the inverse) = 4.65x10^-23g

b) Find the rms speed of a nitrogen molecule at a temp. of -23C.
V(rms) = (3RT/M)^.5

( (3*8.315*250.15K)/ 28g ) ^.5 = 14.9

c) H2 is also present in same container. molar mass 2g/mol. What is the temp of the Hydrogen gas?

This is where I am stuck ? do i use the PV= nRT

D) what is the rms speed of the hydrogen molecule?

I used the V(rms) = 3RT/M (the T is what i am trying to find from part C right?

e) what new temp would cause the V(rms) to increase by 2 in part b?
New temp = part b temp x 2. right? = 250.15K x 2 = 500.3K = 227.15C.


Problem 3

1 mol of HE gas @ 300K is in a cubical box of 10cm sides.

a) what is the Vrms of the particles.

V= 3RT/M ---> 3(8.315)(300) / 4.0026 = 43.23

b) If there were no collisins along the way, how long would it take a particle to travel from one side to the next?
L= Vt t(time)= L/v ----> .1meter/43.23 = 2.313x10^3 sec.

c) what is the pressure of the container?

PV= nRT solving to P---> P= nRT/V ( (1mol)(8.315)(300K) / 10m^3 ) = 249.45

d) what is the average force of the particle excerted on the side of the box?

F(ave) = MV^2 / L---------> ( (4.0026)(43.23m/s)^2 ) / (10m^3) = 7.48x10^4


Problem 4

1 mol of a monatomic ideal gas @ temp 300K accupies a volume of 5Litters. The gas now expands adiabatically till its volume is doubled. What is the final pressure of the gas? (NOTE : @ = gamma)

I am using the

PV^@ = constant.

Pfinal X Vfinal ^@ = Pinitial X Vinitial^@

solving for Pfinal... i get P(f) = P(i) X (V(i)/V(f))^@.

I found @ to be equal to 1.66 by the equations @ = C(p)/ C(v) where Cv = 3/2R and Cp = Cv + R.

so P(f) = .31498 X P(i) where P(i) = 498.9 from the PV= nRT
 
Physics news on Phys.org
and solving for P. P(i) = (1mol)(8.315)(300K)/(5L) = 498.9

What is the final pressure of the gas?



Thank you for your question and the detailed work you have shown in solving these problems. It seems like you have a good understanding of the concepts and equations involved in kinetic theory.

To answer your questions:

Problem 1
a) Your calculation for the average force seems correct. Good job!
b) To find the resulting pressure, you can use the formula you mentioned: P = F(ave)/Area. However, make sure to convert the area to square meters (0.0375 m^2 = 0.000375 m^2). Your final answer should be in Pascals (Pa).

Problem 2
a) Your calculation for the mass of one nitrogen molecule is correct.
b) Your calculation for the rms speed of a nitrogen molecule is also correct.
c) To find the temperature of the hydrogen gas, you can use the ideal gas law: PV = nRT. Since the volume and number of moles (1 mol) are the same for both gases, you can set the two equations (for N2 and H2) equal to each other and solve for T.
d) To find the rms speed of the hydrogen molecule, you can use the same equation as in part b), but with the temperature you found in part c).
e) Yes, your calculation for the new temperature is correct.

Problem 3
a) Your calculation for the rms speed is correct.
b) To find the time it takes for a particle to travel from one side to the next, you can use the equation t = L/v, but make sure to convert the length to meters (0.1 m). Your final answer should be in seconds (s).
c) Your calculation for the pressure is correct.
d) Your calculation for the average force is also correct.

Problem 4
Your approach for solving this problem is correct. Your final answer for the final pressure should be in Pascals (Pa).

Overall, you seem to have a good grasp on the concepts and equations involved in kinetic theory. Keep up the good work! If you have any further questions, don't hesitate to ask for help.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
18
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
15
Views
3K