Practice Questions

Worked Answer:

The density relates mass and volume per atom. The basic idea:

1. Find the mass of one atom from the molar mass.
2. Use the density to find the volume per atom.
3. Use the simple cubic geometry to get the atomic radius from that volume.

1. Mass of one atom

The molar mass is $ M = 56.0 \, \text{g mol}^{-1} = 5.60 \times 10^{-2} \, \text{kg mol}^{-1} $.

Mass of one atom is $$ m_{\text{atom}} = \frac{M}{N_A} \,, $$ where $ N_A \approx 6.02 \times 10^{23} \, \text{mol}^{-1} $.

So $$ m_{\text{atom}} = \frac{5.60 \times 10^{-2}}{6.02 \times 10^{23}} \, \text{kg} \approx 9.3 \times 10^{-26} \, \text{kg}. $$

(Keeping two significant figures is enough for this estimate.)

2. Volume per atom

Density is $ \rho = \frac{\text{mass}}{\text{volume}} $, so the volume per atom is $$ V_{\text{atom}} = \frac{m_{\text{atom}}}{\rho} = \frac{9.3 \times 10^{-26} \, \text{kg}}{8.0 \times 10^{3} \, \text{kg m}^{-3}}. $$

Numerically: $$ V_{\text{atom}} \approx \frac{9.3}{8.0} \times 10^{-26 - 3} \, \text{m}^3 \approx 1.2 \times 10^{-29} \, \text{m}^3. $$

3. Simple cubic geometry and atomic radius

In a simple cubic lattice:

• There is one atom per cube of side length $ a $ (one eighth of an atom at each of the 8 corners).
• Neighbouring atoms touch along the cube edge, so the atomic diameter is equal to the cube side: $ 2r = a $.
• Therefore the volume per atom is the cell volume: $$ V_{\text{atom}} = a^3 = (2r)^3 = 8r^3. $$

So $$ r^3 = \frac{V_{\text{atom}}}{8} \approx \frac{1.2 \times 10^{-29}}{8} \approx 1.5 \times 10^{-30} \, \text{m}^3. $$

Now take the cube root: $$ r = \left( 1.5 \times 10^{-30} \right)^{1/3} \, \text{m}. $$

Compute approximately:

• $ \left( 10^{-30} \right)^{1/3} = 10^{-10} $,
• $ 1.5^{1/3} \approx 1.15 $.

So $$ r \approx 1.15 \times 10^{-10} \, \text{m}. $$

Convert to Ångströms: $$ 1 \, \text{\AA} = 10^{-10} \, \text{m} \; \Rightarrow \; r \approx 1.15 \, \text{\AA} \approx 1.1 \, \text{\AA}. $$

So the best choice is

$$ \boxed{ r \approx 1.1 \, \text{\AA} \text{ (option C)} }. $$

Worked Answer:

The latent heat per mole is the total energy required to separate all the atoms in $ 1 $ mole of the liquid into non-interacting gas atoms.

Each atom has $ z = 12 $ nearest neighbours, but each bond is shared between two atoms, so the number of distinct bonds per mole is $$ N_{\text{bonds}} = \frac{z N_A}{2} . $$

If $ \varepsilon $ is the energy of a single nearest-neighbour bond, then the total energy to break all such bonds in one mole is $$ E_{\text{total}} = N_{\text{bonds}} \varepsilon = \frac{z N_A}{2} \, \varepsilon . $$

This must equal the latent heat per mole (in joules): $$ L_v = E_{\text{total}} = \frac{z N_A}{2} \, \varepsilon . $$

Solving for $ \varepsilon $, $$ \varepsilon = \frac{2 L_v}{z N_A} . $$

1. Convert $ L_v $ to joules per mole: $$ L_v = 8.0 \, \text{kJ mol}^{-1} = 8.0 \times 10^{3} \, \text{J mol}^{-1} . $$
2. Substitute $ L_v = 8.0 \times 10^{3} \, \text{J mol}^{-1} $, $ z = 12 $, and $ N_A = 6.0 \times 10^{23} \, \text{mol}^{-1} $ (approximate value for estimation): $$ \varepsilon = \frac{2 \times 8.0 \times 10^{3}}{12 \times 6.0 \times 10^{23}} \, \text{J} = \frac{16.0 \times 10^{3}}{72.0 \times 10^{23}} \, \text{J} \approx 2.2 \times 10^{-21} \, \text{J} . $$
3. Convert this to electronvolts using $ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} $: $$ \varepsilon = \frac{2.2 \times 10^{-21} \, \text{J}}{1.6 \times 10^{-19} \, \text{J eV}^{-1}} \approx 1.4 \times 10^{-2} \, \text{eV} . $$

So the nearest-neighbour bond energy is $$ \boxed{ \varepsilon \approx 1.4 \times 10^{-2} \, \text{eV} } $$ which corresponds to option B.

Worked Answer:

From kinetic theory, the average translational kinetic energy per molecule is linked to the absolute temperature by $$ \frac{1}{2} m \overline{c^{2}} = \frac{3}{2} k T , $$ where $ m $ is the mass of one molecule, $ \overline{c^{2}} $ is the mean square speed, $ k $ is Boltzmann's constant, and $ T $ is the absolute (Kelvin) temperature.

Rearranging, $$ \overline{c^{2}} = \frac{3kT}{m} \quad\Rightarrow\quad c_{\text{rms}} = \sqrt{\overline{c^{2}}} = \sqrt{\frac{3kT}{m}} . $$

It is often more convenient to write this in terms of molar quantities. Using $ k = \frac{R}{N_A} $ and $ m = \frac{M}{N_A} $ (where $ M $ is the molar mass), $$ c_{\text{rms}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3(R/N_A)T}{M/N_A}} = \sqrt{\frac{3RT}{M}} . $$

So we will use $$ c_{\text{rms}} = \sqrt{\frac{3RT}{M}} . $$

1. Convert temperature to Kelvin

The temperature in Kelvin is $$ T = 27.0^{\circ}\text{C} + 273.15 = 300.15 \, \text{K} \approx 300 \, \text{K}. $$

2. Convert molecular mass from atomic mass units to kg/mol

A molecular mass of $ 20.0 \, \text{u} $ corresponds to a molar mass of $$ M = 20.0 \, \text{g mol}^{-1} = 0.0200 \, \text{kg mol}^{-1}. $$

(Here we have used that the numerical value in u is the same as in $ \text{g mol}^{-1} $, then converted grams to kilograms.)

3. Insert values into the rms speed formula

Take $ R \approx 8.31 \, \text{J mol}^{-1}\text{K}^{-1} $ and $ T \approx 300 \, \text{K} $. Then $$ c_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{ \frac{3 \times 8.31 \, \text{J mol}^{-1}\text{K}^{-1} \times 300 \, \text{K}}{0.0200 \, \text{kg mol}^{-1}} } . $$

First compute the numerator: $$ 3 \times 8.31 \times 300 \approx 3 \times 2493 \approx 7479 \, \text{J mol}^{-1}. $$

Divide by $ M $: $$ \frac{7479}{0.0200} \approx 3.74 \times 10^{5} \, \text{J kg}^{-1}. $$

Now take the square root: $$ c_{\text{rms}} \approx \sqrt{3.7 \times 10^{5}} \, \text{m s}^{-1} \approx 6.1 \times 10^{2} \, \text{m s}^{-1}. $$

So the rms speed is about $$ \boxed{c_{\text{rms}} \approx 6.1 \times 10^{2} \, \text{m s}^{-1}} $$ which corresponds to

$$ \boxed{\text{Option C}}. $$

Common pitfalls

• Forgetting to convert $ 27.0^{\circ}\text{C} $ to Kelvin (using $ T = 27 $ instead of $ T \approx 300 $) gives a speed smaller by a factor of about $ \sqrt{300/27} \approx 3.3 $.
• Treating $ 20.0 \, \text{u} $ as $ 20.0 \, \text{kg mol}^{-1} $ instead of $ 0.0200 \, \text{kg mol}^{-1} $ reduces the speed by a factor of $ \sqrt{1000} \approx 32 $. Both errors lead to answers far from the correct option.

Worked Answer:

For an ideal gas, the equipartition result gives the average energy per molecule as $$ \langle E \rangle = \frac{f}{2} k T , $$ where $ f $ is the number of active quadratic degrees of freedom.

Per mole, the internal energy is $$ U = \frac{f}{2} R T . $$

The molar heat capacity at constant volume is $$ C_v = \left( \frac{\partial U}{\partial T} \right)_V = \frac{f}{2} R . $$

For an ideal gas, the relation between molar heat capacities is $$ C_p = C_v + R . $$

Therefore, $$ \gamma = \frac{C_p}{C_v} = \frac{C_v + R}{C_v} = 1 + \frac{R}{C_v} = 1 + \frac{R}{\frac{f}{2} R} = 1 + \frac{2}{f} . $$

Now substitute $ f = 7.5 $: $$ \gamma = 1 + \frac{2}{7.5} = 1 + 0.266\overline{6} \approx 1.27 . $$

So the correct choice is

$$ \boxed{\gamma \approx 1.27 \text{ (option C)}} . $$

Worked Answer:

First relate the macroscopic van der Waals parameters to the critical temperature, then connect that temperature to a microscopic energy scale using kinetic theory.

1. Compute the critical temperature $ T_c $

We are given $$ T_c = \frac{8a}{27 R b} \, . $$

Insert the numerical values (keeping track of units):

• $ a = 1.0 \, \text{Pa m}^6\text{mol}^{-2} $,
• $ b = 1.0\times 10^{-4} \, \text{m}^3\text{mol}^{-1} $,
• $ R \approx 8.3 \, \text{J mol}^{-1}\text{K}^{-1} = 8.3 \, \text{Pa m}^3\text{mol}^{-1}\text{K}^{-1} $.

Then $$ T_c = \frac{8 \times 1.0 \, \text{Pa m}^6\text{mol}^{-2}} {27 \times 8.3 \, \text{Pa m}^3\text{mol}^{-1}\text{K}^{-1} \times 1.0\times 10^{-4} \, \text{m}^3\text{mol}^{-1}} \, . $$

Check the units:

• Numerator: $ \text{Pa m}^6\text{mol}^{-2} $.
• Denominator: $ \text{Pa m}^3\text{mol}^{-1}\text{K}^{-1} \times \text{m}^3\text{mol}^{-1} = \text{Pa m}^6\text{mol}^{-2}\text{K}^{-1} $.

So $ T_c $ has units of K, as required.

Now evaluate numerically. First the denominator (ignoring units for the moment): $$ 27 \times 8.3 \times 1.0\times 10^{-4} \approx 224.1 \times 10^{-4} = 2.24\times 10^{-2} \, . $$

Thus $$ T_c \approx \frac{8}{2.24\times 10^{-2}} \, \text{K} \approx 3.6\times 10^{2} \, \text{K} \approx 3.6\times 10^{2} \, \text{K} \, . $$

So, to one significant figure, $$ T_c \sim 3.6\times 10^{2} \, \text{K} \, . $$

2. Relate $ T_c $ to a microscopic energy scale

From kinetic theory (derived in the lecture), $$ \text{mean translational kinetic energy per molecule} = \frac{3}{2} k T \, . $$

The problem states that the characteristic interaction energy $ \varepsilon $ is of the same order as this kinetic energy at the critical temperature. So we take $$ \varepsilon \sim \frac{3}{2} k T_c \, . $$

Using $ k \approx 1.4\times 10^{-23} \, \text{J K}^{-1} $ and $ T_c \approx 3.6\times 10^{2} \, \text{K} $: $$ \varepsilon \approx \frac{3}{2} \times 1.4\times 10^{-23} \, \text{J K}^{-1} \times 3.6\times 10^{2} \, \text{K} \, . $$

Compute step by step:

1. $ 1.4\times 10^{-23} \times 3.6\times 10^{2} \approx 5.0\times 10^{-21} \, \text{J} $ (since $ 1.4\times 3.6 \approx 5.0 $ and $ 10^{-23}\times 10^{2} = 10^{-21} $).
2. Multiply by $ \frac{3}{2} \approx 1.5 $: $$ \varepsilon \approx 1.5 \times 5.0\times 10^{-21} \, \text{J} \approx 7.5\times 10^{-21} \, \text{J} \, . $$

So the characteristic interaction energy per molecule is $$ \varepsilon \sim 7\times 10^{-21} \, \text{J} \, . $$

3. Convert $ \varepsilon $ to electronvolts

Using $ 1 \, \text{eV} = 1.6\times 10^{-19} \, \text{J} $, $$ \varepsilon \, (\text{in eV}) = \frac{7.5\times 10^{-21} \, \text{J}} {1.6\times 10^{-19} \, \text{J eV}^{-1}} \approx \frac{7.5}{1.6} \times 10^{-2} \, \text{eV} \, . $$

Now $$ \frac{7.5}{1.6} \approx 4.7 \, , $$ so $$ \varepsilon \approx 4.7\times 10^{-2} \, \text{eV} \, . $$

Rounded to one significant figure in the mantissa, $$ \boxed{ \varepsilon \approx 5\times 10^{-2} \, \text{eV} } \, , $$ which corresponds to option C.

So the correct answer is

$$ \boxed{\text{C) } 5\times 10^{-2} \, \text{eV}} \, . $$

Worked Answer:

First convert the initial temperature to Kelvin: $$ T_1 = 27 \, ^\circ\text{C} \approx 300 \, \text{K}. $$

For a diatomic gas at this temperature, there are three translational and two rotational quadratic degrees of freedom, so $ f = 5 $.

From equipartition, the internal energy per mole is $$ U = \frac{f}{2} R T = \frac{5}{2} R T, $$ so the molar heat capacity at constant volume is $$ C_V = \left( \frac{\partial U}{\partial T} \right)_V = \frac{5}{2} R. $$

Using $ C_P = C_V + R $ for an ideal gas, $$ C_P = \frac{5}{2} R + R = \frac{7}{2} R. $$

Therefore, the heat capacity ratio is $$ \gamma = \frac{C_P}{C_V} = \frac{\frac{7}{2} R}{\frac{5}{2} R} = \frac{7}{5}. $$

For a reversible adiabatic process of an ideal gas, $$ T V^{\gamma - 1} = \text{constant}. $$

Thus, $$ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}. $$

The final volume is one-fifth of the initial: $$ V_2 = \frac{V_1}{5}. $$

So $$ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} = T_1 \left( \frac{V_1}{V_1/5} \right)^{\gamma - 1} = T_1 \, 5^{\gamma - 1}. $$

With $ \gamma = \frac{7}{5} $, we have $$ \gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5}, $$ so $$ T_2 = 300 \, \text{K} \times 5^{2/5}. $$

Estimate $ 5^{2/5} $: $$ 5^{2/5} \approx 1.9 \quad \Rightarrow \quad T_2 \approx 300 \times 1.9 = 570 \, \text{K}. $$

Convert back to Celsius: $$ T_2 \approx 570 \, \text{K} - 273 \, \text{K} \approx 297 \, ^\circ\text{C} \approx 3.0 \times 10^2 \, ^\circ\text{C}. $$

So the closest option is

$$ \boxed{ \text{C) } 3.0 \times 10^2 \, ^\circ\text{C} }. $$

Worked Answer:

Thermal efficiency is defined as $$ \eta = \frac{W}{Q_H} , $$ where $ W $ is the work done per cycle and $ Q_H $ is the heat absorbed from the hot reservoir per cycle.

We are told $$ \eta = 0.300, \qquad W = 630 \, \mathrm{J} . $$

Step 1: Find the heat absorbed from the hot reservoir, $ Q_H $ $$ Q_H = \frac{W}{\eta} = \frac{630 \, \mathrm{J}}{0.300} = 2100 \, \mathrm{J} . $$

This is the magnitude of the heat taken in by the engine from the hot reservoir. From the hot reservoir’s point of view, its heat change per cycle is $$ Q_{\text{hot,res}} = -Q_H = -2100 \, \mathrm{J}, $$ since it loses heat to the engine.

Step 2: Relate the reservoir temperatures using the Carnot efficiency

For a Carnot engine operating between hot and cold reservoirs at absolute temperatures $ T_H $ and $ T_C $, $$ \eta = 1 - \frac{T_C}{T_H} . $$

We are given the cold-reservoir temperature in Celsius: $$ T_C = 21^\circ\mathrm{C} = 21 + 273 = 294 \, \mathrm{K} . $$

Solve for $ T_H $: $$ \eta = 1 - \frac{T_C}{T_H} \quad\Rightarrow\quad 0.300 = 1 - \frac{294 \, \mathrm{K}}{T_H} \quad\Rightarrow\quad \frac{294 \, \mathrm{K}}{T_H} = 1 - 0.300 = 0.700 , $$ so $$ T_H = \frac{294 \, \mathrm{K}}{0.700} = 420 \, \mathrm{K} . $$

Step 3: Entropy change of the hot reservoir

For a reversible heat transfer at constant reservoir temperature, the entropy change is $$ \Delta S = \frac{Q_{\text{rev}}}{T} . $$

For the hot reservoir, $$ \Delta S_{\text{hot}} = \frac{Q_{\text{hot,res}}}{T_H} = \frac{-2100 \, \mathrm{J}}{420 \, \mathrm{K}} = -5.0 \, \mathrm{J \, K^{-1}} . $$

So the hot reservoir loses entropy each cycle.

\[ \boxed{\Delta S_{\text{hot}} = -5.0 \, \mathrm{J \, K^{-1}}} \]

The correct option is

\[ \boxed{\text{A)}\ -5.0 \, \mathrm{J \, K^{-1}}} \]

Worked Answer:

For a constant-volume gas thermometer using an ideal gas, the ideal gas law reduces to $$ P V = n R T \quad \Rightarrow \quad \frac{P}{T} = \text{constant} $$ as long as $ V $ and $ n $ are unchanged. This means $$ \frac{P_1}{T_1} = \frac{P_2}{T_2} \quad \Rightarrow \quad T_2 = T_1 \, \frac{P_2}{P_1} \,. $$

Here:

• Calibration point (triple point of water): $ T_1 = 273.16 \, \text{K} $, $ P_1 = 68.7 \, \text{cmHg} $.
• Unknown sample: $ P_2 = 91.6 \, \text{cmHg} $, $ T_2 = ? $

Notice that we do not need to convert from $\text{cmHg}$ to pascals, because the same non-SI unit is used for both $ P_1 $ and $ P_2 $; it cancels in the ratio.

Compute the ratio $$ \frac{P_2}{P_1} = \frac{91.6}{68.7} \approx 1.33 \,. $$

Now find the absolute temperature $ T_2 $: $$ T_2 = 273.16 \, \text{K} \times \frac{91.6}{68.7} \approx 273.16 \, \text{K} \times 1.33 \approx 364 \, \text{K} \,. $$

Convert this Kelvin temperature to Celsius: $$ T_\text{C} = T_\text{K} - 273.15 \approx 364 - 273 \approx 91^\circ\text{C} \,. $$

So the sample temperature is closest to $ 91^\circ\text{C} $.

$\boxed{\text{Correct answer: C) } 91^\circ\text{C}}$

Worked Answer:

For a crystal, the density is the mass per unit cell divided by the volume per unit cell.

1. Convert lattice parameter to metres

The lattice parameter is

$$ a = 0.287\,\text{nm} = 0.287\times 10^{-9}\,\text{m} = 2.87\times 10^{-10}\,\text{m}. $$

2. Volume of one cubic unit cell

$$ V_{\text{cell}} = a^{3} = \left( 2.87\times 10^{-10}\,\text{m} \right)^{3} \approx 2.36\times 10^{-29}\,\text{m}^{3}. $$

3. Mass of atoms in one BCC unit cell

A BCC unit cell contains 2 atoms.

Molar mass of Fe:

$$ M = 55.8\,\text{g mol}^{-1} = 5.58\times 10^{-2}\,\text{kg mol}^{-1}. $$

Mass of one Fe atom (using Avogadro's number $ N_A \approx 6.02\times 10^{23} \, \text{mol}^{-1} $):

$$ m_{\text{atom}} = \frac{M}{N_A} = \frac{5.58\times 10^{-2}\,\text{kg mol}^{-1}}{6.02\times 10^{23}\,\text{mol}^{-1}} \approx 9.3\times 10^{-26}\,\text{kg}. $$

Mass of one BCC unit cell (2 atoms):

$$ m_{\text{cell}} = 2\,m_{\text{atom}} \approx 2\times 9.3\times 10^{-26}\,\text{kg} \approx 1.86\times 10^{-25}\,\text{kg}. $$

4. Density

$$ \rho = \frac{m_{\text{cell}}}{V_{\text{cell}}} \approx \frac{1.86\times 10^{-25} \, \text{kg}}{2.36\times 10^{-29} \, \text{m}^{3}} \approx 7.9\times 10^{3} \, \text{kg m}^{-3}. $$

To two significant figures,

$$ \boxed{ \rho \approx 7.8\times 10^{3}\,\text{kg m}^{-3} } $$

so the correct option is C

Worked Answer:

First find the tension in the cable, which equals the weight of the load (taking the cable to be massless):

$$ F = mg = 1000\,\text{kg} \times 9.8\,\text{m s}^{-2} = 9.8 \times 10^{3}\,\text{N}. $$

The cross-sectional area of the cable is

$$ A = \pi r^{2}, $$

where the radius is

$$ r = \frac{d}{2} = \frac{12.0\,\text{mm}}{2} = 6.0\,\text{mm} = 6.0 \times 10^{-3}\,\text{m}. $$

So

$$ A = \pi (6.0 \times 10^{-3} \, \text{m})^{2} = \pi \times 3.6 \times 10^{-5} \, \text{m}^{2} \approx 1.13 \times 10^{-4} \, \text{m}^{2}. $$

Young's modulus is

$$ E = 210\,\text{GPa} = 2.10 \times 10^{11}\,\text{Pa}. $$

For a uniform rod under tension,

$$ E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L / L}. $$

Rearranging for the extension $ \Delta L $,

$$ \Delta L = \frac{F L}{A E}. $$

Substitute the numbers:

$$ \Delta L = \frac{(9.8 \times 10^{3} \, \text{N})(15.0 \, \text{m})} {(1.13 \times 10^{-4} \, \text{m}^{2})(2.10 \times 10^{11} \, \text{Pa})}. $$

Compute the denominator:

$$ A E \approx (1.13 \times 10^{-4})(2.10 \times 10^{11}) \approx 2.38 \times 10^{7} \, \text{N}. $$

Compute the numerator:

$$ F L = (9.8 \times 10^{3})(15.0) = 1.47 \times 10^{5} \, \text{N m}. $$

Thus

$$ \Delta L \approx \frac{1.47 \times 10^{5}}{2.38 \times 10^{7}} \, \text{m} \approx 6.2 \times 10^{-3} \, \text{m} = 6.2 \, \text{mm}. $$

The elastic potential energy stored in a linearly elastic cable under constant load is the work done in stretching it from zero extension to $ \Delta L $:

$$ U = \frac{1}{2} F \Delta L. $$

Therefore,

$$ U = \frac{1}{2} \times 9.8 \times 10^{3} \, \text{N} \times 6.2 \times 10^{-3} \, \text{m}. $$

First multiply inside:

$$ F \Delta L \approx (9.8 \times 10^{3})(6.2 \times 10^{-3}) \approx 60.8 \, \text{J}. $$

Then

$$ U \approx \frac{1}{2} \times 60.8 \, \text{J} \approx 30 \, \text{J}. $$

So the elastic energy stored is

$$ \boxed{U \approx 3.0 \times 10^{1} \, \text{J}}, $$

which corresponds to option C.