Conceptual Questions s1

CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY

CONCEPTUAL QUESTIONS ______

1. REASONING AND SOLUTION a. Avogadro's number NA is the number of particles per mole of substance. Therefore, one mole of hydrogen gas (H2) and one mole of oxygen gas (O2) contain the same number (Avogadro's number) of molecules.

b. One mole of a substance has a mass in grams that is equal to the atomic or molecular mass of the substance. The molecular mass of oxygen is greater than the molecular mass of hydrogen. Therefore, one mole of oxygen has more mass than one mole of hydrogen. ______

3. REASONING AND SOLUTION A tightly sealed house has a large ceiling fan that blows air out of the house and into the attic. The fan is turned on, and the owners forget to open any windows or doors. As the fan transports air molecules from the house into the attic, the number of air molecules in the house decreases. Since the house is tightly sealed, the volume of the house remains constant. If the temperature of the air inside the house remains constant, then from the ideal gas law, PV  nRT , the pressure in the house must decrease. The air pressure in the attic, however, increases. The fan must now blow air from a lower pressure region to a higher pressure region. Thus, it becomes harder for the fan to do its job. ______

7. REASONING AND SOLUTION Atmospheric pressure decreases with increasing altitude. Helium filled weather balloons are underinflated when they are launched. As the balloon rises, the pressure exerted on the outside of it decreases. The number of helium molecules in the balloon is fixed. If we assume that the temperature of the atmosphere remains constant, then from the ideal gas law, PV  nRT , we see that the volume of the helium in the balloon will increase, thereby further inflating the balloon. If the balloon is fully inflated when it is launched from earth, it would burst when it reaches an altitude where the expanded volume of the helium is greater than the maximum volume of the balloon. ______

10. REASONING AND SOLUTION The kinetic theory of gases assumes that a gas molecule rebounds with the same speed after colliding with the wall of a container. From the impulse-

momentum theorem, Equation 7.4, we know that Ft  mv f  v 0 , where the magnitude of F is the magnitude of the force on the wall. This implies that in a time interval t, a gas molecule of mass m, moving with velocity v before the collision and –v after the collision, exerts a force of magnitude F  2mv/t on the walls of the container, since the magnitude of v f  v 0 is 2v. If the speed of the gas molecule after the collision is less than that before the collision, the force exerted on the wall of the container will be less than F  2mv/t , since the magnitude of v f  v 0 is then less than 2v. Therefore, the pressure of the gas will be less than that predicted by kinetic theory. ______

11. REASONING AND SOLUTION The relationship between the average kinetic energy per particle in an ideal gas and the Kelvin temperature T of the gas is given by Equation 14.6: 1 2 3 2mvrms  2 kT . If the translational speed of each molecule in an ideal gas is tripled, then the root-mean- square speed for the gas is also tripled. From Equation 14.6, the Kelvin temperature is proportional to the square of the root-mean-square speed. Therefore, the Kelvin temperature will increase by a factor of 9. ______

13. REASONING AND SOLUTION According to Equation 14.7, the internal energy of a sample consisting of n moles of a monatomic ideal gas at Kelvin temperature T is 3 U 2 nRT ; therefore, the internal energy of such an ideal gas depends only on the Kelvin temperature. If the pressure and volume of this sample is changed isothermally, the internal energy of the ideal gas will remain the same. Physically, this means that the experimenter would have to change the pressure and volume in such a way, that the product PV remains the same. This can be verified from the ideal gas law (Equation 14.1), PV  nRT ; if the values of P and V are varied so that the product PV remains constant, then T will remain constant and, from Equation 14.7, the internal energy of the gas remains the same. ______

PROBLEMS ______

2. REASONING AND SOLUTION a. The molecular mass of C55H72MgN4O5 is

55(12.011 u) + 72(1.00794 u) + 24.305 u + 4(14.0067 u) + 5(15.9994 u) = 893.51 u

b. The mass per mol is 893.51 g/mol. The mass of 3.00 mol is

(893.51 g/mol) (3.00 mol) = 2680 g ______

4. REASONING The number of molecules in a known mass of material is the number n of moles of the material times the number NA of molecules per mole (Avogadro's number). We can find the number of moles by dividing the known mass m by the mass per mole.

SOLUTION Using the periodic table on the inside of the text’s back cover, we find that the molecular mass of Tylenol (C8H9NO2) is Molecular mass  8 (12.011 u)  9 (1.00794 u) of Tylenol 1 4 2 4 3 1 4 4 2 4 4 3 Mass of 8 Mass of 9 carbon atoms hydrogen atoms

+ 14.0067 u + 2 (15.9994) =151.165 u 14 2 4 3 1 4 2 4 3 Mass of Mass of 2 nitrogen atom oxygen atoms

The molecular mass of Advil (C13H18O2 ) is

Molecular mass  13 (12.011 u)  18 (1.00794 u) + 2 (15.9994) = 206.285 u of Advil 1 4 4 2 4 4 3 1 4 4 2 4 4 3 1 4 2 4 3 Mass of 13 Mass of 18 Mass of 2 carbon atoms hydrogen atoms oxygen atoms

a. Therefore, the number of molecules of pain reliever in the standard dose of Tylenol is

m  Number of molecules = n N  N A  A Mass per mole 

3 325  10 g  23 1 21  6.022  10 mol  1.29  10 151.165 g/mol 

b. Similarly, the number of molecules of pain reliever in the standard dose of Advil is

m  Number of molecules = n N  N A  A Mass per mole 

 2.00  10 g  23 1 20  6.022  10 mol  5.84  10 206.285 g/mol  ______

9. REASONING We can use the ideal gas law, Equation 14.1 (PV = nRT) to find the number of moles of helium in the Goodyear blimp, since the pressure, volume, and temperature are known. Once the number of moles are known, we can find the mass of helium in the blimp.

SOLUTION The number n of moles of helium in the blimp is, according to Equation 14.1,

5 3 PV (1.1  10 Pa)(5400 m ) n    2.55  105 mol RT [8.31 J/(mol  K)](280 K) According to the periodic table on the inside of the text’s back cover, the atomic mass of helium is 4.002 60 u. Therefore, the mass per mole is 4.002 60 g/mol. The mass m of helium in the blimp is, then,

51 kg  3 m 2.55  10 mol 4.002 60 g/mol   1.0  10 kg 1000 g  ______

12. REASONING The number N of molecules in an ideal gas is given by Equation 14.2 as N PV/ kT , where P, V, and T are the absolute pressure, volume, and Kelvin temperature of the gas, and k is Boltzmann’s constant. The ratio of the number of air molecules in the two rooms is equal to the ratio of the ideal gas laws for the air in each room.

SOLUTION The number of air molecules in rooms A and B is

PA V A P B V B NAand N B  (14.2) kTA kT B

Dividing NA by NB, and noting that PA = PB and VA = VB, we see that

PA V A N kT T 19 + 273 K A A  B   1.10 P V NBB B T A 8.0  273 K

kTB ______

20. REASONING a. The volume V of gas in the tank is related to the Kelvin temperature T and absolute pressure P of the gas by Equation 14.1, V nRT/ P , where n is the number of moles of gas and R is the universal gas constant.

b. The mass of chlorine gas that has leaked out of the tank is equal to the initial mass (11.0 g) minus the mass remaining in the tank at the later time. The mass at the later time is equal to the mass per mole (in g/mol) of chlorine times the number n of moles remaining in the tank. The number of moles of gas remaining can be found from the ideal gas law, n PV/ RT (Equation 14.1)

SOLUTION a. The number n of moles of chlorine originally in the tank is equal to the mass of chlorine (11.0 g) divided by the mass per mole of chlorine (70.9 g/mol), so n = (11.0 g)/(70.9 g/mol). The temperature of the gas is 82 C, which, when converted to the Kelvin temperature scale, is (82 + 273 )K. The volume of the tank is 11.0 g    8.31 J/ mol K   273  82 K  nRT 70.9 g /mol     (14.1) V    8.17  104 m 3 P 5.60 105 Pa b. The mass mCl of chlorine gas that has leaked out of the tank is mCl = 11.0 g – mR, where mR is the mass of the gas remaining in the tank. The remaining mass, in turn, is equal to the number nR of moles of gas times the mass per mole (70.9 g/mol) of chlorine. Thus,

mCl11.0 g  m R = 11.0 g  n R  70.9 g/mol

The number of remaining moles of gas can be found from the ideal gas law, nR  PV/ RT , so that PV  m 11.0 g    70.9 g/mol Cl RT  3.80 105 Pa 8.17  10 4 m 3  11.0 g  70.9 g/mol  2.3 g 8.31 J/ mol K   273  31 K  ______

26. REASONING AND SOLUTION The volume of the cylinder is V = AL where A is the cross- sectional area of the piston and L is the length. We know P V = P V so that the new 1 1 2 2 pressure, P can be found. We have 2

P2 = P1(V1/V2) = P1A1L1/A2L2 = P1L1/L2 (since A1 = A2)

P = (1.01  105 Pa)(L/2L) = 5.05  104 Pa 2

The force on the piston and spring is, therefore, 4 2 F = P2A = (5.05  10 Pa) (0.0500 m) = 397 N

The spring constant is k = F/x so k = F/x = (397 N)/(0.200 m) = 1.98 103 N/m ______

PV 2 N( KE )  28. REASONING AND SOLUTION From Equation 14.5 3  , we obtain

3PV 3(4.5 105 Pa)(8.5  10 3 m 3 ) KE    4.8  10–21 J 2N 23 1  2 (2.0 mol)(6.022 10 mol )  ______

35. REASONING AND SOLUTION Since the argon atom has a kinetic energy equal to its average translational kinetic energy,

1 3 mv 2  kT 2 rms 2 (1)

Using conservation of mechanical energy, we have Eearth's surface = Ehighest altitude, or 1 mv 2  mgh 2 rms , where the gravitational potential energy near the earth's surface has been chosen to be zero. Using Equation (1), this statement of energy conservation becomes 3 kT  mgh . Solving for h gives 2 3kT h  (2) 2mg

In order to use Equation (2), we must find the mass of the argon atom:

39.948 g/mol m   6.634  10–23 g = 6.634  10 –26 kg 6.022  10 23 mol –1

Substituting into Equation (2) gives

3(1.38  10–23 J/K)(295 K) h   9400 m 2(6.634  10 –26 kg)(9.80 m/s 2 ) ______

37. REASONING AND SOLUTION The average force exerted by one electron on the screen is, from the impulse-momentum theorem (Equation 7.4), F  p/ t  mv/ t . Therefore, in a time t , N electrons exert an average force F  Nmv/ t  (N / t)mv . Since the pressure on the screen is the average force per unit area (Equation 10.19), we have

F (N / t)mv P   A A (6.2  1016 electrons/s)(9.11  10–31 kg)(8.4  107 m/s – 0 m/s)   4.0  101 Pa 1.2  10–7 m2 ______