General Physics II

Be sure and do these; do not just wait and watch me do them in class!

Q21.13 An ideal gas is contained in a vessel at 300 K. If the temperature is increased to 900 K,

(a) by what factor does the rms speed of each molecule change?

The temperature has increased by a factor of 3. The rms speed goes like the square root of the temperature so we have increased the rms speed by SQRT(3)=1.732

(b) by what factor does the pressure in the vessel change?

The Ideal Gas Law still holds, P V = n R T. If the volume remains constant, an increase in temperature by a factor of 3 means an increase in pressure by the same factor of 3.

SQRT[] means "Square Root of" because it is easier for me to write SQRT than to go to a graphics editor and construct a Square Root sign.

21.*. Find the rms speed of nitrogen molecules under standard conditions, 0.0^oC and 1.00 atm pressure. Recall that 1 mole of any gas occupies a volume of 22.4 liters under standard conditions.

Eq 21.7, on page 590, tells us

T = 0^oC = 273 K

R = 8.31 J / mol-K

M = 28 g/mole = 28 x 10^{- 3} kg/mole

v_rms = SQRT[ 3 (8.31 J/mol-K)(273 K)/(28 x 10^{- 3} kg/mole)]

v_rms = SQRT [2.43 x 10⁵ (J/mol-K)(K)/(kg/mole)]

v_rms = SQRT [2.43 x 10⁵ (J/kg)(N-m/J)(kg-m/s²/N)]

v_rms = SQRT [2.43 x 10⁵ (m²/s²)]

v_rms = 493 m/s

Be careful with the units. They're important! They're not just an afterthought.

21.3 In a 30-s interval, 500 hailstones strike a glass window of area 0.60 m² at an angle of 45^o to the window surface. Each hailstone has a mass of 5.0 g (0.005 kg) and a speed of 8.0 m/s. If the collisions are elastic, find the average force and pressure on the window.

v_x = v cos 45^o

v_x = (8.0 m/s)(0.707) = 5.66 m/s

p_x = m v_x = (0.005 kg)(5.66 m/s) = 2.83 x 10^{- 2} kg-m/s

p_x = 2 p_x = 5.66 x 10^{- 2} kg-m/s

F = p_xtot/t

F = (500)(5.66 x 10^{- 2} kg-m/s)/30 s

F = 0.943 kg-m/s²

F = 0.943 N

P = F/A

P = 0.943 N / 0.60 m²

P = 1.57 N/m2

P = 1.57 Pa

21.** Calculate the rms speed of an H₂ molecule at 250^oC.

Of course, this is basically the same as problem 21.1 above. But it is always good to work similiar problems -- just to see that the idea is the same even with different numbers. This time

T = 250^oC = 523 K

M = 2 g/mole = 2 x 10^{- 3} kg/mole

As with 21.1, we begin with Eq 21.7, from page 590,

R = 8.31 J / mol-K

v_rms = SQRT[ 3 (8.31 J/mol-K)(523 K)/(2 x 10^{- 3} kg/mole)]

v_rms = SQRT [6.52 x 10⁶ (J/mol-K)(K)/(kg/mole)]

v_rms = SQRT [6.52 x 10⁶ (J/kg)(N-m/J)(kg-m/s²/N)]

v_rms = SQRT [6.52 x 10⁶ (m²/s²)]

v_rms = 2 550 m/s

21.8 If the rms speed of a helium atom at room temperature is 1350 m/s, what is the rms speed of an oxygen (O₂) molecule at this temperature? (The molar mass of O₂ is 32 and the molar mass of He is 4.)

This problem, too, begins with Eq 21.7,

v_rmsO/v_rmsH = SQRT[3 R T/M_O] / SQRT[3 R T/M_H]

v_rmsO/v_rmsH = SQRT[ (3 R T/M_O) / (3 R T/M_H) ]

v_rmsO/v_rmsH = SQRT[ M_H/M_O ]

v_rmsO/v_rmsH = SQRT[ 4/16 ]

v_rmsO/v_rmsH = SQRT[ 1/4 ]

v_rmsO/v_rmsH = 1/2

v_rmsO/v_rmsH = 0.5

v_rmsO = 0.5 v_rmsH

v_rmsO = 0.5 (1350 m/s)

v_rmsO = 675 m/s

21.*** One mole of xenon gas at 20.0^oC occupies 0.0224 m³. What is the pressure exerted by the Xe atoms on the walls of a container?

P V = n R T

P = n R T / V

P = (1 mole) (8.31 J/mole-K) (293 K) / 0.0224 m³

P = 1.09 x 10⁵ (J/m³)

Now, what in the world is a J/m³?

P = 1.09 x 10⁵ (J/m³) [ N-m / J ]

P = 1.09 x 10⁵ N/m²

P = 1.09 x 10⁵ Pa

P = 1.09 x 10⁵ Pa [1 atm / 1.013 x 10⁵ Pa]

P = 0.029 atm

21.13 Calculate the change in internal energy of 3.0 moles of helium gas when its temperature is increased by 2.0 K.

For an ideal gas, we would have

U = (3/2) n R T

U = (3/2) (3 moles) (8.31 J/mole-K) (2.0 K)

U = 74.8 J

While helium behaves very nearly as an ideal gas, we can do this calculation for the actual behavior of helium.

U = n C_v T

From Table 21.2 on page 647 of the text, we find that, for helium

c_v = 12.5 J/mole-K

U = (3 moles)(12.5 J/mole-K)(2.0 K)

U = 75.0 J

21.**** One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process acquiring 500 J of heat. It then undergoes an isobaric process losing the same amount of heat. Determine (a) the new temperature of the gas and (b) the work done on the gas.

[[ Note: Part of the earlier solution posted on the web was wrong.]]

For the isovolumetric (constant volume) process, from A to B, no work is done,

W_AB = 0

Q = U = 500 J

For the isobaric (constant pressure), process, from B to C, the work is

W_BC = P V

We know the initial temperature, T_A = 300 K. To find W_BC we will need the temperatures T_B and T_C at states B and C so we can find P V for the expression above.

Q_AB = n C_V T = 500 J

(n) ((3/2) R) (T) = 500 J

(1 mole) ((3/2) 8.31 J/mole-K) (T) = 500 J

( 12.5 J / K ) T = 500 J

T = (500/12.5) K

T = 40 K

T_B = 300 K + 40 K

T_B = 340 K

Q_BC = n C_PT

- 500 J = (n) ((5/2) R) (T)

- 500 J = (1 mole) ((5/2) 8.31 J/mole-K) (T)

- 500 J = ( 20.8 J / K ) T

T = - (500/20.8) K

T = - 24.1 K

T_C = 340 K - 24.1 K

T_C = 315.9 K

W_tot = W_AB + W_BC

W_AB = 0

W_BC = P V

Q_tot = W_tot + U_tot

Q_tot = 500 J + ( - 500 J) = 0

W_tot = - U_tot

U = n C_V T

U = (n) ((3/2) R) (T_tot)

T = 40.0 K - 24.1 K

T = 15.9 K

U = (n) ((3/2) R) (T_tot)

U = (1 mole)(12.5 J/mole-K)(15.9 K)

U = 199 J

W_tot = - U_tot

W = - 199 J

[[ This final answer is the same as was posted earlier but part of the earlier solution was wrong. ]]

21.25 Two moles of an ideal gas ( = 1.40) expands slowly and adiabatically from a pressure of 5.0 atm and a volume of 12.0 liters to a final volume of 30.0 liters.

(a) What is the final pressure of the gas?

"Adiabatically" means

P_f = (5.0 atm) (12.0 L/30.0 L)^1.4

P_f = (5.0 atm) (0.4)^1.4

P_f = (5.0 atm) (0.2773)

P_f = 1.386 atm

(b) What are the initial and final temperatures?

Again (or "as usual"?!?!?), we will find the Ideal Gas Law very useful;

P V = n R T

T = P V / n R

T_i= P_i V_i / n R

T_i = (5.0 atm) (12.0 L)/[(2.0 moles) (8.31 J/mole-K)]

Be careful! While that equation is true, it is also clumsy or cumberson since we have such strange or mixed units. Since we have pressure in atmospheres and volume in liters, we will be far better served to use R in units of L-atm instead of joules! Again (and again and again and again!), watch the units! Units are not an "add-on"; they're important -- even vital!

T_i= (5.0 atm) (12.0 L)/[(2.0 moles) (0.0821 L-atm/mole-K)]

T_i = 365.4 K

T_f = P_f V_f / n R

T_f = (1.386 atm) (30.0 L)/[(2.0 moles) (0.0821 L-atm/mole-K)]

T_f = 253.2 K

21.***** Four liters of a diatomic ideal gas ( = 1.40) confined to a cylinder are put through a closed cycle. The gas is initially at 1.0 atm and at 300 K. First, its pressure is tripled under constant volume. It then expands adiabatically to its original pressure and finally is compressed isobarically to its original volume.

(z) What is its initial pressure?
P V = n R T
P = n R T / V

(a) Draw a PV diagram of this cycle.

Here is a rough sketch:

(b) Determine the volume at the end of the adiabatic expansion.

As we move along the adiabat,

or

or

(c) Find the temperature of the gas at the start of the adiabatic expansion

At state A,
P_A V_A = n R T_A
n = P_A V_A / R T_AA

At state B, the start of the adiabatic expansion,
P_B V_B = n R T_BB
T_B = P_B V_B / n R

T_B = P_B V_B /[(P_A V_A / R T_A) R]

T_B = T_A [ P_B V_BB /(P_A V_A) ]

T_B = T_A (P_B/P_A) (V_B/V_A)

[ or, we could have gotten this about as easily by noting
P V = n R T
P V / T = n R

P_A V_A/T_A = P_B V_B/T_B

T_B = T_A (P_B/P_A) (V_B/V_A)

Having done this both ways, this way seems noticeably easier. But, of course, the result is the same.]
T_B = T_A (P_B/P_A) (V_B/V_A)
Of course,
V_B = V_A
T_B = 300 K (3 atm/1 atm) (1)

T_B = 300 K (3)

T_B = 900 K

(d) Find the temperature at the end of the cycle.

We know the pressure and volume for state C, the end of the adiabatic expansion,
P_C = 1 atm
V_C = 8.77 L

P V = n R T

P V / T = const

P_A V_A / T_A = P_C V_C / T_C

T_C = T_A (P_C/P_A) (V_C/VA)

T_C = 300 K (1 atm/1 atm) (8.77 L/4 L)

T_C = 300 K (1) (2.19)

T_C = 657.5 K

(e) What was the net work done for this cycle?

Look at the work for each piece and then sum those;
W_AB = 0
W_BC = - U

T = T_f - T_i = T_C - T_B

T = 657.5 K - 900 K

T = - 242.5

U = n C_V T

U = n [ (5/2) R ] [ - 242.5 K ]

How many moles do we have? Go back to state A and find out;
P V = n R T
n = [ P V/T ] [1/R]

n = [ 1 atm - 4 L / 300 K] [1/R]

U = [ 1 atm - 4 L / 300 K] [1/R] [ (5/2) R ] [ - 242.5 K ]

U = [ 4 atm - L / 300 ] [ (5/2)] [ - 242.5 ]

U = - 8.08 L-atm

We have done this unit coversion before. We could just go back and get it. But it's worth doing from "scratch", too;
U = - 8.08 L-atm [0.001 m3/L] [1.013 x 10⁵ (N/m²)/atm]
U = - 819 J

W_BC = + 819 J

W_CA = P V

W_CA = (1 atm) (4 L - 8.77 L)

W_CA = - 4.77 L-atm

W_CA = - 4.77 L-atm [0.001 m³/L] [1.013 x 10⁵ (N/m²)/atm]

W_CA = - 483 J

W_tot = W_AB + W_BC + W_CA

W_tot = 0 + 819 J - 483 J

W_tot = 336 J

While I have labeled this W_tot, as in the total work, we could also call this the net work and label it W_net.

21.****** A 5.0-liter vessel contains 0.125 mole of an ideal gas at 1.50 atm. What is the average translational kinetic energy of a single molecule?

<KE> = (1/2) m <v²>

<KE> = (1/2) m <v²> = (1/2) k T

So, first, we must find the temperature T

P V = n R T

T = P V / n R

T = (1.5 atm) (5.0 L) / [(0.125 mole) (0.0821 L-atm/mole-K)]

T = 731 K

<KE> = (1/2) k T = (1/2) (1.38 x 10^{- 23} J/K) (731 K)

<KE> = 5.04 x 10^{- 21} J

21.33 Consider 2.0 moles of an ideal diatomic gas. Find the total heat capacity at constant volume and at constant pressure if

(a) the molecules rotate but do not vibrate and
C_V = (5/2) R
C_tot = n C_V = 5 R

(b) the molecules rotate and vibrate.
C_V = (7/2) R
C_tot = n C_V = 7 R

Kinetic Theory of Gases

Homework Solutions