F1 (a) Since the ideal gas equation of state is PV = nrt, we can equate the right-hand sides of both these equations (i.e. with PV = 2 3 N ε tran )and write: nrt = 2 3 N ε tran = 2 3 nn A ε tran i.e. T = 2 3 nn A nr ε tran = 2 N A 3 R ε tran = 2 3k ε tran This provides a definition of the macroscopic temperature T in terms of the average of the microscopic translational kinetic energy. (b) We just rearrange the equation from (a): ε tran = 2 3 kt = 1. 5 1. 381 10 23 JK 1 295 K = 6.11 10 21 J Note that if we know the temperature, we do not need the number of moles (or atoms) to calculate the mean kinetic energy per atom; we would need that information to calculate the total internal energy.
(c) For five degrees of freedom, the equipartition of energy theorem states that the internal energy will be 5 2 NkT or 5 2 nrt. So E int = 2 5 5 mol 8.314 J K 1 mol 1 295K = 3. 07 10 4 J (d) The specific heat at constant pressure is given by C P = C V + R = 2 5 R + R = 2 7 R = 2 7 8.314 J K 1 mol 1 = 29.10 J K 1 mol 1
F2 (a) n(v) v represents the number of atoms (or molecules) in the gas which have speeds in the small range v to (v + v), so it gives information on how the population of atoms is distributed among the range of possible speeds. (The area under a graph of n(v) against v, over the entire range of speeds, will just be the total number of atoms in the gas.) (b) The mass of one mole of molecular hydrogen is 2.021g, which means that the mass of an individual hydrogen molecule (which is what m in the equation represents) is 2.021g/(6.02 10 23 ) = 3.36 10 24 1g = 3.36 10 27 1kg. Then v rms = 3 1.381 10 23 JK 1 250 K 3.36 10 27 kg = 3. 08 10 6 m 2 s 2 = 1. 76 10 3 ms 1 (c) Since their root-mean-squared speeds are the same, but their masses are different, their temperatures must be different as well. We can rewrite the equation as T = m 3k v 2 rms, so the molecule with the larger mass will have the higher temperature. The oxygen will be hotter. (Section 3 gives a fuller discussion of the Maxwell Boltzmann distribution.)
F3 (a) There are two differences between this equation and the ideal gas equation of state. First, instead of P we have the quantity [P + (a/v m2 )]. This correction to the ideal gas behaviour represents the effect of intermolecular interactions. (See Subsection 4.3 for a more detailed explanation.) The second difference is replacing the volume V m by the term (V m b). This is a correction for the finite size of the molecules (which are points in the ideal gas model) and accounts for the excluded volume arising from the volumes of all the other molecules. (See Subsection 4.2 for more detail.) (b) We rearrange the equation slightly and substitute in the given values: T = (P + av m 2 )(V m b) R = 5.00 10 6 Pa + 2.30 10 2 Nm 4 mol 2 (7.50 10 4 m 3 mol 1 ) 2 (7.50 10 4 2.70 10 5 )m 3 mol 1 8.314 J K 1 mol 1 = 435K (Make sure you understand the unit conversions above.)
(c) Again, we need to rearrange the equation to solve for P. P + i.e. P = a V m 2 = RT V m b RT V m b a V 2 m Now we substitute in the given values (remembering that V m (per mole) is five times the given value of V): 8.314 J K 1 mol 1 438K P = 2.25 10 3 m 3 mol 1 2.70 10 5 m 3 mol 1 2.30 10 2 Nm 4 mol 2 ( 2.25 10 3 m 3 mol 1 ) 2 P = 3.64 103 J mol 1 2.22 10 3 m 3 mol 1 2.30 10 2 Nm 4 mol 2 5.06 10 6 m 6 mol 2 = 1. 64 106 Jm 3 4.55 10 3 Nm 2 =1. 64 10 6 Pa Again, make sure you understand the unit conversions.
Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to Ready to study? in Subsection 1.3. Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Closing items. If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it here.
R1 (a) The average speed 1v1 is calculated by taking the ratio of the total distance travelled to the total time taken: 1v1 = (5 + 5)1m/101s = 11m1s 1 (b) The average velocity is calculated by taking the ratio of the net displacement to the total time taken for the displacement. Here the net displacement is zero and so 1v1 = 101 1m/101s with 1 1v1 1 = 01m1s 1 The average velocity is zero because the net displacement is zero since the particle returns to its starting point. (c) The average kinetic energy is m 1v 2 1 /2 = 31kg (11m1s 1 ) 2 /2 = 1.51J. (d) For the average magnitude of momentum, we take the product of the mass with the average speed: 1p1 = m 1v1 = 31kg 11m1s 1 = 31kg1m1s 1
(e) Taking the momentum to be positive in the initial direction of motion: momentum before the turnaround = p 1 = 31kg1m1s 1 in the initial direction momentum after turnaround = p 2 = 31kg1m1s 1 in the initial direction momentum change for particle is (1p 2 p 1 ) = 61kg1m1s 1 in the initial direction. (f) According to Newton s second law of motion, the average force exerted on the wall is equal to the rate of change of momentum, so F = (1p 2 p 1 )/ t = (6/ t)1kg1m1s 2 in the initial direction. Consult the relevant terms in the Glossary for further details.
T1 (a) Using the ideal gas law, PV = nrt, we find: n = PV ( 2.00 10 5 Pa) 2.00 m 3 = RT 8.314 J K 1 mol 1 310 K n = 4.00 105 Pa m 3 2.58 10 3 J mol 1 = 4 10 5 Nm 2.58 10 3 1 = 155moles N m mol (b) The mass (in grams) of one mole of any substance is equal to its relative molecular mass, so for molecular hydrogen this is 2.021g. Thus, the system contains 2.021g 155 = 3131g = 0.3131kg. The mass density is then ρ = M 0.313 kg = = 0.157 kg m 3 V 2.00 m3 (c) We have PV = constant, so if V increases by a factor of 2, P must decrease by a factor of 2. Thus, the new pressure is 1.00 10 5 1Pa.
T2 (a) Using Equation 1 PV = nrt with n = 11mol and R = 8.3141J1K 1 1mol 1, we find the volume of the gas V = nrt P = 1mol 8.314 J K 1 mol 1 300 K 1. 00 10 5 Nm 2 =2.49 10 2 m 3 (Eqn 1) (b) The volume occupied by the molecules themselves is: N 4 A 3 πr3 = 6.02 1023 mol 1 4 π (2 10 10 m) 3 = 2 10 5 m 3 mol 1 3 = 2 10 5 m 3 for n = 1mol So, 1 mole of molecules occupies 2 10 5 1m 3, which is less than 0.1% of the total gas volume. Therefore Assumption 5 is justified.
(c) The number of molecules per unit volume is nn A V = 1mol 6.02 1023 mol 1 2.49 10 2 m 3 = 2.42 10 25 m 3 (d) If molecules have an average separation d, then each volume d0 3 contains on average one molecule. Total gas volume V = nn A d0 3 so d 3 = 2.49 10 2 m 3 1mol 6.02 10 23 mol 1 3 2.49 10 and d = 2 m = 3. 46 10 9 m = 3.46 nm 6.02 10 23 The ratio d r = 3. 46 10 9 m 0.2 10 9 m = 17 So, the average molecular separation is 17 times the molecular radius.
Alternatively, we could have obtained the value for d0 3 from the answer to part (c). If the number of molecules per volume is x then d 3 = 1 x = 1m 3 2.42 10 25
T3 From Equation 9, T = 2 N A 3 R ε tran (Eqn 9) for both gases ε tran = 3 2 8.314 J K 1 mol 1 300 K 6.02 10 23 mol 1 = 6.21 10 21 J To calculate the speed, use Equation 12, v rms = 3kT m For helium, m is 6.6 10 27 1kg, so v rms = 3 1.38 10 23 JK 1 300 K 6.6 10 27 = 1.37 km s 1 kg For argon, m = 6.6 10 26 1kg, so v rms = 4341m1s 1 Notice that although the average kinetic energies are the same, the helium atoms travel faster because they are lighter.
T4 11eV corresponds to an energy, in SI units, of 1.60 10 19 1J. From Equation 10 For the monatomic gas ε tran = 2 3 kt (Eqn 10) we have ε tran = 3 2 kt so T = 2 ε tran 3k so T = 2 ε tran 2 1. 6 10 19 J = 3k 3 1.38 10 23 JK 1 =7.72 103 K For hydrogen molecules (m = 3.3 10 27 1kg), at this temperature, Equation 12 gives: v rms = 3kT m = 3 1.38 10 23 JK 1 7.73 10 3 K 3.3 10 27 kg = 9.85 km s 1
T5 For the two samples each containing one mole we have energy added = E int = 3 2 = 5 2 Nk ( T ) monatomic Nk ( T ) diatomic where N = N A 11mol in each case, so ( T ) diatomic = 3 5 ( T ) monatomic So the monatomic system reaches the higher temperature. This is because in the monatomic molecule the temperature corresponds only to the amount of translational kinetic energy in the system; in the case of the diatomic molecule, the energy is shared out between translational and rotational kinetic energy.
T6 Using Equations 18 and 22, C V = q 2 N Ak = q R 2 (Eqn 18) C P C V = R (Eqn 22) we have the generalization: C P = q 2 R + R = q + 2 R 2
T7 The values for C V, C P, (C P C V ) and γ, as given in Table 2, are determined from Equations 18, 22, 23 and 24. C V = q 2 N Ak = q R (Eqn 18) 2 C P C V = R (Eqn 22) γ = C P C V = 1 + R = 1 + R 2 = 1 + C V q 2 R q C P = 1 + q 2 (Eqn 23) R (Eqn 24) Table 23See Answer T7. q C V = qr/2 C P = (1 + q/2)r (C P C V ) = R γ = C P /C V = 1 + 2/q 3 3R/2 5R/2 R 5/3 5 5R/2 7R/2 R 7/5 7 7R/2 9R/2 R 9/7
T8 If we substitute the figures from Question T2 into Equation 28 (i.e. an ideal gas at T = 3001K and at a pressure of 1.00 10 5 1Pa. ) 1 λ = (Eqn 28) 4 2πr 2 n ρ we find: 1 λ = 4 2 π (2 10 10 ) 2 m 2 2.42 10 25 = 58 nm 3 m This is equivalent to about 145 molecular diameters and is about 17 times the intermolecular separation.
T9 Yes, they are consistent. The first estimate is that there will be another molecule within about 8.75 diameters in some direction, while the second value is estimating how far the molecule will have to travel in a particular direction before colliding with another molecule. The second value depends on the size of the targets but the first does not. For the mean free path, we are essentially asking how long a cylinder with twice the diameter of a molecule must be to have the same volume as the cube that, on average, contains one molecule.
T10 The exponential function in Equation 30 32 m f (v) v = 4π v 2πkT 2 exp ( mv 2 2kT ) v (Eqn 30) is dimensionless and has no units. The units of v 2 are (m 2 1s 2 ), so we just need to work out the units of the factor 32 m. 2πkT We know kt has the units of energy (k = 1.38 10 23 1J1K 1 ), so its units are kg1m 2 1s 2. The ratio then has units (kg/kg1m 2 1s 2 ) = m 2 1s 2, which is equivalent to 1/v 2. This, taken to the power of 3/2, produces a factor with the dimensions of 1/v 3. Finally then, the dimensions of the terms on the right-hand side apart from v are those of 1/v, which must be the dimensions of f1(v), as required.
T11 We know that heating the gas adds translational kinetic energy and so molecular speeds, on average, must increase1 1so the peak must shift to a higher speed. We also know that the effect of increasing T is to reduce the quantity A(T) and to reduce the exponentially decaying factor in 32 m f (v) = 4π v 2 exp( mv 2 2kT) = AT ( )v 2 exp( mv 2 2kT) 2πkT The curve for the higher temperature must rise less rapidly at low speeds (as A(T) is less) and fall less rapidly at high speeds (as the exponential factor is less negative). This implies that the curve is broadened, i.e. there are a wider range of speeds. The area under the whole curve is unity (as the probability for a molecule to have a speed somewhere within the full range is unity). So if the area under the curve stays constant but it is broadened it follows that the curve must reach a lower peak.
T12 The average speed for the five molecules is: v = v 1 + v 2 + v 3 + v 4 + v 5 = 1 + 2 + 2 + 3 + 4 = 12 = 2.40 (in the arbitrary units of the question) 5 5 5 For this very simple distribution the most probable speed, v prob, is 2. The root-mean-squared (rms) speed is given by Equation 6: where v rms = v 2 v 2 = v 1 2 + v 2 2 + v 3 2 + v 4 2 + v 5 2 5 = 12 + 2 2 + 2 2 + 3 2 + 4 2 5 = 34 5 = 6.80 We find v rms = 6.80 = 2.61, which confirms the claim,v rms > 1v1 > v prob. It can be seen from the averaging processes above that the evaluation of v rms gives additional weight to the higher values of v as compared to the lower values of v (by squaring the numbers); in contrast, the calculation of 1v1 treats large and small speeds equally. It follows that v rms > 1v1 is true for any distribution (unless all molecules have the same speed).
T13 We use Equations 32, 33 and 12, the most probable speed v prob = the average speed v = the root-mean-squared speed v rms = with the helium mass substituted: v prob = v = v rms = 2kT m 8kT πm 3kT m 2kT m = 2 1.38 10 23 J K 1 300 K 6.65 10 27 kg 8kT πm = 8 1.38 10 23 J K 1 300 K π 6.65 10 27 kg 3kT m = 3 1.38 10 23 J K 1 300 K 6.65 10 27 kg = 1.12 km s 1 = 1.26 km s 1 = 1.37 km s 1 (Eqn 32) (Eqn 33) (Eqn 12)
T14 To be dimensionally consistent, the terms within any set of brackets must have the same units. Thus, the units of b must be m 3 1mol 1 and a/v m 2 must be in Pa (= N1m 2 ). Table 13Values of van der Waals constants a and b for various gases. a/10 1 1m 6 1Pa1mol 2 b/10 5 1m 3 1mol 1 helium (He) 0.035 2.4 nitrogen (N 2 ) 1.41 3.91 oxygen (O 2 ) 1.38 3.18 xenon (Xe) 4.25 5.10 So a must have the units of P V m 2 and since V m 2 has units of m 6 1mol 2, this means that the units of a are N1m 2 m 6 1mol 2, or N1m 4 1mol 2. Therefore the units given in Table 1 are correct.
T15 Perhaps surprisingly, the answer is that there is no effect at all1 1whatever the details of the force! For simplicity, consider an attractive force which comes into play only very near the wall. This force will accelerate the molecules into the collision but decelerate them after the collision. The net change of momentum of the molecules will be the same as in the no force situation and it will be as if no force were operating.