tido_29
Feb4-04, 05:14 PM
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 dont 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
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