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course phy 232
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?****
KE_trans = (3/2)nRT, so then I get this: (3/2)(10mol)(240K)(8.31) = 30 kJ
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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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since a diatomic gas has more degrees of rotation.( not a single sphere, so it has rotational KE as well, but not 3 degrees of rotation because it has symmetry about one axis), so additionally to the (3/2) another 2 halves will consist of its TOTAL KE, so the equation is this: (5/2)(10)(240)(8.31) = 50 kJ
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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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I know its monatomic and an ideal gas, so I know (3/2)nRT = KE_tran. I know the KE = 5000 J, so I can solve for T, so (3/2)(10)(T)(8.31)=5000, so change in T must be 40.1 Kelvin
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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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using the same process as above, except this time, taking in account of the rotational KE, which would add (2/2) to the system, so now I get: 5000 = (10)(T)(8.31)(5/2), so T = 24 Kelvin
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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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since the gas is diatomic, the total KE is (5/2)nRT, so therefore the energy required is: (5/2)(8.31)(100*change in temp)(45) = 93.5 kJ
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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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throwing pressure into this changes it up a bit. I know PV=nRT, and I know that KEtotal = (5/2)nRT I need to relate the two equations using T and P in my equation. I know V, n, AND r stay constant, and I know my new P. so I can solve for the required T, which is 133.7 K, throwing that into the total KE equation for diatomic molecules, I get (5/2)(45)(8.31)(133.7) = 125 kJ
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How would this result change if the gas was monatomic?
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my temperature increase would be the same, but KE change would be less since more energy could be in translation and not rotation, so less total energy would be required. So (3/2)(45)(8.31)(133.7) = 75 kJ
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`q003. What is the average translational KE per particle of a gas at temperature 300 K?
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my total KE is (3/2)nRT, so that is: (3/2)(45)(8.31)(300) = 169 kJ. To find the KE per particle, I'll have to divide by Avogadro's number and 45, which is 6.02E23. So my average KE per particle is:
6.21E-21 J
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What therefore would be the rms speed of particles of mass 1.66 * 10^-27 kg?
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I know the KE per particle, and I can use the KE=.5mv^2, with the given mass above, I can solve for it’s speed. So I have (6.21E-21 = (.5)(1.66E-27 kg)v^2, so v= 2765 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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Just plug in 9m into the equation, and I get v=912 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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the translational KE is (3/2)nRT, or (3/2)kT, where k is R/N, which is 6.21E-21 J, so I know that KE = .5mv^2, so now I have 6.21E-21 = .5(32*1.66E-27)(300)v^2, so v is 27.9 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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KErot is (2/3) of KEtrans, so that is (R/N)T, so then I get 1.86E-18 J
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@& You previously found the average translational KE of a molecule.
The average rotational KE will be 2/3 of this.
Not sure where you got R / N * T ... wait, maybe I do ... do you mean R / N_A * T? That would be right, but your result would be on the order of 10^-21 J.*@
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&#Good work. See my notes and let me know if you have questions. &#