CHEM 161 Exam 3A Study Guidance: Gas Laws, Thermodynamics, and Calorimetry

Q1. A carbon dioxide sample weighing 44.0 g occupies 32.68 L at 65°C and 645 torr. What is its volume at STP?

Background

Topic: Gas Laws (Combined Gas Law)

This question tests your ability to use the combined gas law to relate the volume of a gas under different conditions of temperature and pressure.

Key formula:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Where:

• $P_1$ = initial pressure (645 torr)

• $V_1$ = initial volume (32.68 L)

• $T_1$ = initial temperature (65°C, convert to K)

• $P_2$ = final pressure (STP = 1 atm = 760 torr)

• $V_2$ = final volume (what you're solving for)

• $T_2$ = final temperature (STP = 0°C = 273 K)

Step-by-Step Guidance

1. Convert all units to the proper form: pressure in torr or atm, temperature in Kelvin.

2. Set up the combined gas law equation with the known values.

3. Rearrange the equation to solve for $V_2$.

4. Plug in the values, but stop before calculating the final result.

Try solving on your own before revealing the answer!

Final Answer: 34.3 L

$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}$

Plugging in the values and converting units gives the final volume at STP.

Q2. Which of the following states the first law of thermodynamics (conservation of energy)?

Background

Topic: Thermodynamics

This question tests your understanding of the first law of thermodynamics, which deals with energy conservation in a system.

Key Terms:

• First Law of Thermodynamics: Energy cannot be created or destroyed, only transferred or transformed.

• $\Delta E_{system} = -\Delta E_{surroundings}$

Step-by-Step Guidance

1. Review each option and identify which one expresses energy conservation.

2. Recall that the first law relates the change in energy of the system to the surroundings.

3. Eliminate options that do not directly relate to energy conservation.

Try solving on your own before revealing the answer!

Final Answer: B. $\Delta E_{system} = -\Delta E_{surroundings}$

This equation directly states the conservation of energy between system and surroundings.

Q3. The kinetic-molecular theory predicts that pressure rises as the temperature of a gas increases because ________.

Background

Topic: Kinetic-Molecular Theory

This question tests your understanding of how temperature affects gas pressure according to kinetic-molecular theory.

Key Concepts:

• As temperature increases, the average kinetic energy of gas molecules increases.

• Pressure is caused by collisions of gas molecules with container walls.

Step-by-Step Guidance

1. Recall what happens to molecular motion as temperature increases.

2. Consider how this affects the frequency and energy of collisions with the container walls.

3. Identify which answer choices reflect these effects.

Try solving on your own before revealing the answer!

Final Answer: D. Both the gas molecules collide more frequently and more energetically with the wall

Higher temperature increases both collision frequency and energy, raising pressure.

Q4. If 16.7 kJ of energy is absorbed by a 225-g sample of benzene at 20.0°C, what is its final temperature? (c= 1.74 J g-1°C-1)

Background

Topic: Calorimetry

This question tests your ability to use the specific heat formula to calculate temperature changes.

Key formula:

$q = m c \Delta T$

Where:

• $q$ = heat absorbed (convert kJ to J)

• $m$ = mass (225 g)

• $c$ = specific heat (1.74 J g-1°C-1)

• $\Delta T$ = change in temperature

Step-by-Step Guidance

1. Convert the energy absorbed from kJ to J.

2. Set up the equation $q = m c \Delta T$ and solve for $\Delta T$.

3. Add $\Delta T$ to the initial temperature to find the final temperature.

4. Check significant figures and units.

Try solving on your own before revealing the answer!

Final Answer: 62.7°C

$\Delta T = \frac{q}{m c}$, then $T_{final} = T_{initial} + \Delta T$

Plugging in the values gives the final temperature.

Q5. A balloon full of hydrogen is lowered to the bottom of a lake that is 10 m deep. If the temperature of the gas does not change, what will happen to the density of the gas?

Background

Topic: Gas Laws (Boyle's Law, Density)

This question tests your understanding of how pressure affects gas density at constant temperature.

Key Concepts:

• Density ($\rho$) is mass per unit volume.

• At constant temperature, increasing pressure decreases volume, increasing density.

Step-by-Step Guidance

1. Recall Boyle's Law: $P_1 V_1 = P_2 V_2$ at constant temperature.

2. As pressure increases, volume decreases.

3. Since mass stays the same, density increases as volume decreases.

Try solving on your own before revealing the answer!

Final Answer: D. Density of hydrogen will increase

Higher pressure compresses the gas, increasing its density.

Q6. “The pressure of an ideal gas is inversely proportional to its volume at constant temperature and number of moles” is a statement of ________ Law.

Background

Topic: Gas Laws

This question tests your knowledge of the different gas laws and their relationships.

Key formula:

$P V = \text{constant}$ (Boyle's Law)

Step-by-Step Guidance

1. Recall which gas law relates pressure and volume inversely.

2. Match the description to the correct law.

3. Eliminate options that do not fit the relationship described.

Try solving on your own before revealing the answer!

Final Answer: B. Boyle’s Law

Boyle's Law states that pressure and volume are inversely related at constant temperature.

Q7. Which of the following gases has a density of 1.2 g/L in a vessel of 3.0 L, at 0.78 atm and 25°C?

Background

Topic: Gas Laws and Density

This question tests your ability to use the ideal gas law and density formula to identify a gas.

Key formula:

$\rho = \frac{m}{V}$

$PV = nRT$

$n = \frac{m}{M}$

Where:

• $\rho$ = density (1.2 g/L)

• $m$ = mass

• $V$ = volume (3.0 L)

• $P$ = pressure (0.78 atm)

• $T$ = temperature (25°C = 298 K)

• $R$ = 0.0821 L·atm/mol·K

• $M$ = molar mass

Step-by-Step Guidance

1. Calculate the mass of gas using density and volume.

2. Use the ideal gas law to solve for the number of moles.

3. Calculate the molar mass using $M = \frac{m}{n}$.

4. Compare the calculated molar mass to the options given.

Try solving on your own before revealing the answer!

Final Answer: C. O2

The calculated molar mass matches that of oxygen gas.

Q8. Which of the following is a correct formation reaction?

Background

Topic: Thermochemistry (Formation Reactions)

This question tests your understanding of standard formation reactions, which form one mole of a compound from its elements in their standard states.

Key Concepts:

• Formation reaction: One mole of a compound is formed from its elements in their standard states.

Step-by-Step Guidance

1. Identify which option forms one mole of a compound from elements in their standard states.

2. Check that the reactants are in their standard states and the product is one mole.

3. Eliminate options that do not meet these criteria.

Try solving on your own before revealing the answer!

Final Answer: D. C (graphite) + 4 H (g) → CH4 (g)

This reaction forms one mole of methane from its elements in their standard states.

Q1 (Short Answer). Calculate the enthalpy of reaction for the formation of one mol solid aluminum chloride from its constituent elements using the following equations:

Background

Topic: Thermochemistry (Hess's Law)

This question tests your ability to use Hess's Law to calculate the enthalpy of formation from given reactions.

Key formula:

Hess's Law: $\Delta H_{rxn} = \sum \Delta H$ (for manipulated equations)

Step-by-Step Guidance

1. Write the target reaction: formation of 1 mol AlCl3 (s) from Al (s), Cl2 (g), and H2 (g).

2. Identify how to combine and manipulate the given equations to achieve the target reaction.

3. Adjust coefficients and reverse equations as needed, keeping track of enthalpy changes.

4. Sum the enthalpy changes for the manipulated equations, but stop before the final calculation.

Try solving on your own before revealing the answer!

Final Answer: -704 kJ

After manipulating and summing the enthalpy changes, the enthalpy of formation is -704 kJ.

Q2 (Short Answer). A gas mixture of N2 and CO2 has a total pressure of 8.00 atm and contains 12.5 mol of gas. How many moles of CO2 are in the mixture if the partial pressure of N2 is 3.69 atm?

Background

Topic: Partial Pressures (Dalton's Law)

This question tests your ability to use Dalton's Law of Partial Pressures to find the mole fraction and moles of a component in a gas mixture.

Key formula:

$P_{total} = P_{N_2} + P_{CO_2}$

$X_{CO_2} = \frac{P_{CO_2}}{P_{total}}$

$n_{CO_2} = X_{CO_2} \times n_{total}$

Step-by-Step Guidance

1. Calculate the partial pressure of CO2 by subtracting the partial pressure of N2 from the total pressure.

2. Find the mole fraction of CO2 using its partial pressure and the total pressure.

3. Multiply the mole fraction by the total number of moles to find moles of CO2.

4. Set up the calculation, but stop before computing the final value.

Try solving on your own before revealing the answer!

Final Answer: 7.13 mol CO2

Using Dalton's Law and mole fraction, the moles of CO2 are calculated.

Q3 (Short Answer). When 15.3 g of sodium nitrate was dissolved in water in a bomb calorimeter, the temperature fell from 25.00 °C to 21.56 °C. If the heat capacity of the solution and the calorimeter is 1071 J/°C, what is the enthalpy change for the process when 1 mol of sodium nitrate dissolves in water?

Background

Topic: Calorimetry

This question tests your ability to calculate enthalpy change using calorimetry data and scale it to per mole.

Key formula:

$q = C \Delta T$

$\Delta H = \frac{q}{n}$

Where:

• $C$ = heat capacity (1071 J/°C)

• $\Delta T$ = temperature change

• $n$ = moles of sodium nitrate

Step-by-Step Guidance

1. Calculate the temperature change ($\Delta T$).

2. Calculate the heat absorbed or released ($q$) using $q = C \Delta T$.

3. Find the number of moles of sodium nitrate dissolved.

4. Set up the calculation for enthalpy change per mole, but stop before the final computation.

Try solving on your own before revealing the answer!

Final Answer: +25.6 kJ/mol

The enthalpy change per mole is calculated using calorimetry data.

Q4 (Short Answer). Calculate the ΔH°rxn for the following reaction. (ΔH°f [AsH3(g)] = 66.4 kJ/mol; ΔH°f [H3AsO4(aq)] = −904.6 kJ/mol; ΔH°f [H2O(l)] = −285.8 kJ/mol)

Background

Topic: Thermochemistry (Standard Enthalpy of Formation)

This question tests your ability to use standard enthalpies of formation to calculate the enthalpy change for a reaction.

Key formula:

$\Delta H^\circ_{rxn} = \sum \Delta H^\circ_f (products) - \sum \Delta H^\circ_f (reactants)$

Step-by-Step Guidance

1. Write the reaction and list the enthalpy of formation values for each compound.

2. Multiply each enthalpy of formation by the stoichiometric coefficient.

3. Sum the enthalpy values for products and reactants.

4. Set up the calculation for ΔH°rxn, but stop before the final computation.

Try solving on your own before revealing the answer!

Final Answer: +2030 kJ

ΔH°rxn is calculated using the enthalpy of formation values and stoichiometry.

Q5 (Short Answer). Calculate both the ideal and non-ideal volume taken up by 0.4891 mol of N2 at 125 atm and 28.5°C. (a = 1.390 L2 atm mol-2; b = 0.0391 L mol-1). Comment on the differences/similarities in values (why are they close or not close, use kinetic molecular theory in your description).

Background

Topic: Real Gases (Van der Waals Equation)

This question tests your ability to calculate gas volume using both the ideal gas law and the van der Waals equation for real gases.

Key formulas:

Ideal Gas Law: $PV = nRT$

Van der Waals Equation: $\left(P + \frac{a n^2}{V^2}\right)(V - n b) = nRT$

Step-by-Step Guidance

1. Calculate the ideal volume using $V = \frac{nRT}{P}$.

2. Set up the van der Waals equation with the given values for a and b.

3. Compare the two volumes and consider why they might differ, referencing kinetic molecular theory.

4. Stop before solving the van der Waals equation for V.

Try solving on your own before revealing the answer!

Final Answer: Ideal: 0.0978 L; Non-ideal: 0.0892 L

The non-ideal volume is less due to intermolecular forces and finite molecular size, as explained by kinetic molecular theory.

Q6 (Short Answer). Which gaseous element diffuses 3.98 times as fast as O2?

Background

Topic: Gas Laws (Graham's Law of Diffusion)

This question tests your ability to use Graham's Law to compare rates of diffusion based on molar mass.

Key formula:

$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$

Step-by-Step Guidance

1. Set up Graham's Law with the given rate ratio.

2. Plug in the molar mass of O2 and solve for the unknown molar mass.

3. Identify which element matches the calculated molar mass.

4. Stop before identifying the element.

Try solving on your own before revealing the answer!

Final Answer: Hydrogen (H2)

Hydrogen's low molar mass allows it to diffuse much faster than O2.

Q7 (Short Answer). A gas is confined to a cylinder under constant pressure of 2.0 atm. When the gas undergoes a particular chemical reaction, it absorbs 824 J of heat from its surroundings and the volume of the gas changes from 6.20 L to 3.00 L. What are the values of enthalpy, work, and energy change for this process? Is the work done on the system or surroundings?

Background

Topic: Thermodynamics (Enthalpy, Work, Internal Energy)

This question tests your ability to calculate work, enthalpy, and energy change for a process at constant pressure.

Key formulas:

$w = -P \Delta V$

$\Delta H = q_p$

$\Delta E = q + w$

Step-by-Step Guidance

1. Calculate the change in volume ($\Delta V$).

2. Calculate the work done using $w = -P \Delta V$ (convert atm·L to J).

3. Enthalpy change is equal to the heat absorbed at constant pressure ($\Delta H = q_p$).

4. Calculate the internal energy change ($\Delta E = q + w$), but stop before the final calculation.

Try solving on your own before revealing the answer!

Final Answer: ΔH = 824 J; w = +646 J; ΔE = 1470 J; Work is done on the system

Work is positive because the volume decreases, so surroundings do work on the system.

Q8 (Short Answer). An aerosol can contains gas under a pressure of 4.50 atm at 20.0°C. If the can is left on a hot, sandy beach, the pressure of the gas increases to 4.80 atm. What is the beach temperature (°C)?

Background

Topic: Gas Laws (Gay-Lussac's Law)

This question tests your ability to relate pressure and temperature for a gas at constant volume.

Key formula:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Where:

• $P_1$ = initial pressure (4.50 atm)

• $T_1$ = initial temperature (20.0°C = 293 K)

• $P_2$ = final pressure (4.80 atm)

• $T_2$ = final temperature (what you're solving for)

Step-by-Step Guidance

1. Convert the initial temperature to Kelvin.

2. Set up the equation $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ and solve for $T_2$.

3. Convert $T_2$ from Kelvin back to Celsius.

4. Stop before calculating the final temperature.

Try solving on your own before revealing the answer!

Final Answer: 313 K (40°C)

The beach temperature is calculated using Gay-Lussac's Law.