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course Phy 242
2-16 12I'm not sure if this has been submitted?????
110202
By considering the momentum change for a single collision and time interval between collisions we determine that a particle of mass m bouncing back and forth at speed v, with velocity always parallel to the axis, between the ends of a cylinder of length L, exerts average force
F_ave = m v^2 / (L)
on each end, and that the average pressure on either end of the cylinder is therefore
P_ave = m v^2 / (L * A_cs),
where A_cs is the cross-sectional area of the cylinder. Since L * A_cs is the volume V of the cylinder we get
P_ave = m v^2 / V
so that
P_ave * V = m v^2.
Since 1/2 m v^2 is the KE of the particle we get
P_ave * V = 2 * KE.
If there are N such particles we get
P_ave * V = N * 2 * KE.
If there are a large number of such particles randomly distributed throughout the cylinder, the collisions occur almost continuously, and we can replace P_ave with just P.
If, realistically, the particles collide with one another their velocities become randomly distributed in space, velocities become distributed according to a probability distribution and our equation becomes
P V = 2/3 N * KE_ave,
where KE_ave = 1/2 v_rms^2 and the symbol v_rms stands the square root of the mean of the squared velocities (called the RMS velocity, where 'RMS' stands for ""Root Mean Square').
N particles comprise n = N / N_A moles, where N_A = 6.02 * 10^-23 is the number of particles in a mole (called Avagodro's Number).
Experiments indicate that for an ideal gas,
P V = n R T,
where n is the number of moles and R the gas constant R = 8.31 Joules / (mole * Kelvin).
Thus
n R T = 2/3 N * KE_ave (for true particles, which consist of point masses),
and since N * KE_ave is the total kinetic energy of the particles,
n R T = 2/3 KE_total (for true particles)
and
KE_total = 3/2 n R T (for true particles).
For each of the 3 possible independent directions of motion in 3-dimensional space, we thus have
KE_df = 1/2 n R T,
where the df in KE_df stands for 'degree of freedom'. Each particle being free to move in 3 dimensions, we say that it has 3 degrees of freedom.
If the 'particles' of the gas are in fact diatomic molecules, consisting of two atoms, then they can also possess rotational KE. A diatomic 'particle' can rotate about any axis except the central axis through the centers of the two atoms. There are three independent directions in space, two of which are perpendicular to and hence independent of the direction of the central axis. So there are two independent directions about which a diatomic molecule can rotate.
A large number of such molecules, colliding elastically with one another and with the molecules in the walls of a container, will distribute their kinetic energy equally among the 3 directions of space and the two independent axes about which each can rotate. Each direction of space and each independent possible axis of rotation consists of a degree of freedom, the particle being free to move in 3 dimensions of space and to rotate about two independent axes.
Thus the total kinetic energy of a diatomic gas is distributed among 5 independent degrees of freedom.
The three directions of space correspond to 'translational' degrees of freedom, since they correspond to a translation of position. The two rotational axes correspond to 'rotational' degrees of freedom.
With three degrees of translational and two degrees of rotational freedom, then, the total KE is divided into 5 parts, with translational KE taking up 3, and rotational KE taking up 2 of these 5 parts. It follows that
KE_df = 1/5 KE_total (for diatomic molecules)
and thus
KE_translational = 3/5 KE_total (for diatomic molecules)
KE_rotational = 2/5 KE_total (for diatomic molecules).
Only the translational kinetic energy has a net effect on pressure, since only translational motion moves the particles toward and away from the walls in the collisions that create pressure.
Recall that KE_df = 1/2 n R T, so for any gas, we have
KE_translational = 3/2 n R T.
For a diatomic gas, we also have
KE_rotational = 2/2 n R T (diatomic gas)
with the result that
KE_total = 5/2 n R T (diatomic gas).
`q001. Suppose we have 10 moles of an ideal monatomic gas (a monatomic gas consists of single, unjoined atoms, and so consists of point masses only). If we are to increase the temperature of this gas by 240 degrees Kelvin, without expansion, how much energy do we need to put into the gas?
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KE_needed= (3/2)*(10 mol)*(8.31 (J/(mol*K)))*(240 K)
= 29,916 Joules
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To increase the temperature of 10 moles of ideal diatomic gas (a gas consisting of pairs of atoms), without expansion, how much energy do we need to put into the gas?
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KE_needed = [(3/2)*n*R*T] + [(2/2)*n*R*T]
=(5/2)*nRT
=(5/2)*(10 mol)*(8.31 (J/(mol*K)))*(240 K)
=49,860 Joules
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If we put 5000 Joules of thermal energy into 10 moles of ideal monatomic gas, without expansion, by how much will the temperature change?
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KE_needed= (3/2)*(10 mol)*(8.31 (J/(mol*K)))*`dT, so plugging in 5,000J for KE
`dT = (2/3)*(5,000J)/((10 mol)*(8.31 (J/(mol*K))
= approx 40 K
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If we put 5000 Joules of thermal energy into 10 moles of ideal diatomic gas, without expansion, by how much will the temperature change?
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KE_needed= (5/2)*(10 mol)*(8.31 (J/(mol*K)))*`dT, so plugging in 5,000J for KE
`dT = (2/5)*(5,000J)/((10 mol)*(8.31 (J/(mol*K))
= approx 24 K
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`q002. If we have a cubic meter of gas at atmospheric pressure and temperature 273 K, then that gas consists of about 45 moles. It doesn't matter whether it's monatomic or diatomic.
Suppose now that the gas is diatomic.
If we want to heat a cubic meter of the gas from 273 K to 373 K, without changing its volume, how much thermal energy is required?
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Thermal energy needed…
(5/2)*(45 mol)*(8.31 (J/(mol*K)))*`dT
= (5/2)*(45 mol)*(8.31 (J/(mol*K)))*100 K
= 93,487.5 Joules
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If we want to heat the gas, starting at 273 K and 1 atmosphere of pressure, at constant volume, until its pressure has increased by 50 000 N / m^2 (i.e., 50 000 Pa or 50 kPa), how much thermal energy will it take?
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Using PV = nRT, where V, n, and R are constant
(P_1)/(T_1) = (P_2)/(T_2), plugging in our values
= (100 kPa)/(273 K) = (150 kPa)/(T_2), solving for T_2
T_2 = 273 K*(150kPa / 100kPa)
= 409.5K, giving us a `dT = 136.5K
Now using following equation we can find Thermal Energy
(5/2)*(45 mol)*(8.31 (J/(mol*K)))*`dT, plugging in our `dT
(5/2)*(45 mol)*(8.31 (J/(mol*K)))*136.5 K
Thermal Energy needed = 127,610.4375 J
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How would this result change if the gas was monatomic?
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We would calculate using formula
(3/2)*(45 mol)*(8.31 (J/(mol*K)))*`dT
Instead of
(5/2)*(45 mol)*(8.31 (J/(mol*K)))*`dT
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`q003. What is the average translational KE per particle of a gas at temperature 300 K?
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KE_trans/# of particles = ((3/2)nRT/N), where N = moles*Avagrado’s #
So we can reduce term n/N
n/N = n/(n*N) = (1/N)
So KE_trans_particle = (3/2)*(RT/N)
Solving in original form
((3/2)*(45 m)*(8.31 (J/(mol*K))*300K)/((45 m)*( 6.02 * 10^-23))
= 168,277.5J / 2.709^(-21)
= 2.06545224E14
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@& To get from Joules / mole to Joules / particle you would divide by the number of particles in a mole.
However there are 6.02 * 10^23 particles in a mole. The exponent is 23, not -23.
You end up with roughly 6 * 10^-21 Joules per particle.*@
What therefore would be the rms speed of particles of mass 1.66 * 10^-27 kg?
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KE_ave_particle_rms = (1/2)*m*v_rms^2, at 300K
KE_trans_particle = (3/2)*(RT/N) = (3/2)*(R/N)*T
= (3/2)*(1.38039867*10^-23 J/(part.*K))*300K
= approx 6.2*10^-21J
(1/2)*m*v_rms^2 = KE_ave
v_rms = `sqrt(2KE_ave/m)
= `sqrt((2*6.2*10^-21)/ 1.66 * 10^-27 kg)
=`sqrt((12.4*10^-21J)/(1.66*10^-27 kg))
= approx `sqrt(7,469,879(m^2/s^2))
= approx 2,733 m/s
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What would be the rms speed of particles whose mass is 9 times as great as in the preceding?
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Using the same equation (1/2)*m*v_rms^2 = KE_ave using a mass 9 times as great
v_rms = `sqrt(2KE_ave/(m*9))
= `sqrt((2*6.2*10^-21)/ (1.66 * 10^-27 kg)*9)
=`sqrt((12.4*10^-21J)/(1.66*10^-27 kg)*9)
= approx `sqrt(829,968(m^2/s^2))
= approx 911 m/s
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Diatomic oxygen molecules have mass about 32 times as great as 1.66 * 10^-27 kg. What is the rms speed of oxygen molecules in a gas at 300 K?
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KE_ave_particle_rms = (1/2)*m*v_rms^2, at 300K
KE_trans_particle = (5/2)*(RT/N) = (5/2)*(R/N)*T
= (5/2)*(1.38039867*10^-23 J/(part.*K))*300K
= approx 1.035*10^-20J
(1/2)*m*v_rms^2 = KE_ave
v_rms = `sqrt(2KE_ave/m)
= `sqrt((2.07*10^-20J)/ (53.12 * 10^-27 kg))
= approx `sqrt(389,683(m^2/s^2))
= approx 624 m/s
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What is the average rotational KE of oxygen molecules in a gas at 300 K?
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KE_rot = ((2/2) nRT/(3/2) nRT)) = (2/3) KE_trans
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The effect of an elastic collisions between a particle and a receding wall is to reduce its speed. If the wall of a container is free to recede, then collisions with confined gas particles accelerate the wall, causing it to recede and reduce the speed of the particles. This causes the average KE of the particles to decrease, hence reducing the temperature of the gas.
If the wall is under pressure from the outside, it takes work to move it, and the work is done at the expense of the KE of the particles.
It is possible to add thermal energy to to the gas during the process, possibly mitigating the reduction of the KE of the particles (reducing the loss of temperature), possibly balancing the energy (hence keeping the temperature constant), perhaps even exceeding the energy required to move the wall (hence increasing the temperature).
The bottom line:
Any time a gas expands, it has to do work against pressure. Unless balanced by the input of thermal energy, the temperature of the gas will therefore decrease with expansion.
`q004. When we heat a gas at constant pressure, it expands, and some of the thermal energy added must do the work of expansion.
If a gas changes volume against constant pressure, then the thermal energy required of the gas is 2/2 n R `dT, which is conveniently equivalent to the thermal energy required for two additional degrees of freedom (there aren't two additional degrees of freedom, but the energy required is the same as if there were).
If the cubic meter of the diatomic gas in question 2 is heated from 273 K to 373 K at constant pressure, how much energy is required?
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Need: (3/2) nR`dT KE_trans
(2/2) nR`dT KE_rot
(2/2) nR`dT `dW of expansion
Total needed = (7/2) nR`dT
= (7/2)*(45m)*(8.31(J/(mol*K))*100K
= 130,882.5 J
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How much of that energy goes into changing the KE of the molecules, and how much into the work of expansion?
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KE of the molecules = (5/7)
work of expansion = (2/7)
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How would these results change if the gas was monatomic?
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No rotational KE needed so…
KE of the molecules = (3/5)
work of expansion = (2/5)
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&#Good responses. See my notes and let me know if you have questions. &#