Physics with Calculus: Thermodynamics and Ideal Gas Law Study Guide

Q1. What is the coefficient of performance of cooling, $K_c$, for the heat pump used as a refrigerator illustrated on the pV diagram below? The gas is diatomic.

Background

Topic: Thermodynamics – Heat Pumps and Refrigerators

This question tests your understanding of the coefficient of performance (COP) for refrigerators, which measures how efficiently a refrigerator moves heat from a cold reservoir to a hot reservoir.

Key Terms and Formulas:

• $K_c$ (Coefficient of Performance for cooling): $K_c = \frac{Q_c}{W}$

• $Q_c$ = heat removed from the cold reservoir

• $W$ = work input to the refrigerator

Step-by-Step Guidance

1. Identify the heat removed from the cold reservoir ($Q_c$) and the work input ($W$) from the pV diagram and the thermodynamic cycle.

2. Recall that for a refrigerator, $K_c = \frac{Q_c}{W}$, where $Q_c$ is the heat extracted from the cold side and $W$ is the net work done.

3. Analyze the cycle on the pV diagram to determine which segments correspond to heat extraction and work input.

4. Set up the expressions for $Q_c$ and $W$ using the first law of thermodynamics and properties of diatomic gases.

Try solving on your own before revealing the answer!

Final Answer: 5

The coefficient of performance for cooling, $K_c$, is 5. This means the refrigerator removes 5 units of heat from the cold reservoir for every unit of work input.

Q2. What is the pressure of 2 mol of ideal gas in a volume of 5 m3 and with a temperature of 800 K? Use $R=10$ J/mol·K.

Background

Topic: Ideal Gas Law

This question tests your ability to use the ideal gas law to calculate the pressure of a gas given the number of moles, volume, and temperature.

Key Terms and Formulas:

• Ideal Gas Law: $PV = nRT$

• $P$ = pressure (Pa)

• $V$ = volume (m3)

• $n$ = number of moles

• $R$ = gas constant (J/mol·K)

• $T$ = temperature (K)

Step-by-Step Guidance

1. Write down the known values: $n = 2$ mol, $V = 5$ m3, $T = 800$ K, $R = 10$ J/mol·K.

2. Plug these values into the ideal gas law: $PV = nRT$.

3. Rearrange the formula to solve for pressure: $P = \frac{nRT}{V}$.

4. Set up the calculation for $P$ using the values above.

Try solving on your own before revealing the answer!

Final Answer: 3200 Pa

$P = \frac{(2)(10)(800)}{5} = 3200$ Pa. The pressure of the gas is 3200 Pa.

Q3. Suppose two identical blocks have the following initial thermal energies: $E_{Ai} = 32$ J and $E_{Bi} = 54$ J. The blocks are thermally connected so that they can exchange heat. Once the blocks reach thermal equilibrium (in an isolated chamber), what is the final energy of block A, $E_{Af}$, and the amount of heat exchanged, $Q$, respectively?

Background

Topic: Thermal Equilibrium and Heat Exchange

This question tests your understanding of energy conservation and heat transfer between two objects reaching thermal equilibrium.

Key Terms and Formulas:

• Thermal equilibrium: Both blocks reach the same final energy.

• Conservation of energy: Total energy before = total energy after.

• Heat exchanged: $Q = |E_{Af} - E_{Ai}|$

Step-by-Step Guidance

1. Calculate the total initial energy: $E_{Ai} + E_{Bi}$.

2. Since the blocks are identical and isolated, the final energies will be equal: $E_{Af} = E_{Bf}$.

3. Set up the equation: $2E_{Af} = E_{Ai} + E_{Bi}$.

4. Express the heat exchanged as the difference between initial and final energy for block A.

Try solving on your own before revealing the answer!

Final Answer: $E_{Af} = 43$ J, $Q = 11$ J

Each block ends up with 43 J of energy, and 11 J of heat is exchanged from block B to block A.

Q4. A bicycle tire is inflated to a gauge pressure of 11 atm at a temperature of 300 K. While a man rides the bicycle, the temperature lowers to 250 K. Assuming the volume does not change, find the gauge pressure of the air in the tire at this lower temperature. Use atmospheric pressure of $p_0 = 1$ atm.

Background

Topic: Ideal Gas Law – Pressure and Temperature Relationship

This question tests your understanding of how gas pressure changes with temperature at constant volume.

Key Terms and Formulas:

• Ideal Gas Law: $PV = nRT$

• At constant volume: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$

• Gauge pressure: $P_{gauge} = P_{total} - p_0$

Step-by-Step Guidance

1. Write down the initial and final temperatures: $T_1 = 300$ K, $T_2 = 250$ K.

2. Write down the initial gauge pressure: $P_{gauge,1} = 11$ atm.

3. Calculate the initial total pressure: $P_{total,1} = P_{gauge,1} + p_0$.

4. Set up the ratio $\frac{P_{total,1}}{T_1} = \frac{P_{total,2}}{T_2}$ and solve for $P_{total,2}$.

5. Subtract atmospheric pressure to find the new gauge pressure.

Try solving on your own before revealing the answer!

Final Answer: 9 atm

The gauge pressure at 250 K is 9 atm. The pressure decreases as the temperature drops, assuming constant volume.

Q5. What is the change in internal (thermal) energy $\Delta E_{th}$ of $n=5$ mol of monatomic ideal gas when its temperature is decreased by 12 K? Use $R=10$ J/mol·K.

Background

Topic: Internal Energy of Ideal Gases

This question tests your understanding of how the internal energy of a monatomic ideal gas changes with temperature.

Key Terms and Formulas:

• Internal energy change: $\Delta E_{th} = n C_v \Delta T$

• For monatomic ideal gas: $C_v = \frac{3}{2}R$

• $n$ = number of moles

• $R$ = gas constant

• $\Delta T$ = change in temperature

Step-by-Step Guidance

1. Write down the values: $n = 5$ mol, $R = 10$ J/mol·K, $\Delta T = -12$ K (since temperature decreases).

2. Calculate $C_v$ for a monatomic gas: $C_v = \frac{3}{2}R$.

3. Set up the formula: $\Delta E_{th} = n C_v \Delta T$.

4. Plug in the values to set up the calculation.

Try solving on your own before revealing the answer!

Final Answer: -900 J

The internal energy decreases by 900 J. The negative sign indicates a loss of thermal energy.

Q6. What is a non-zero quantity?

Background

Topic: Thermodynamic Processes

This question tests your understanding of which quantities are non-zero for different thermodynamic processes (cycle, adiabatic, isochoric, isothermal).

Key Terms and Formulas:

• Cycle: $\Delta E_{th} = 0$ for a complete cycle

• Adiabatic: $Q = 0$

• Isochoric: $W_{tot} = 0$

• Isothermal: $Q \neq 0$, $\Delta E_{th} = 0$

Step-by-Step Guidance

1. Review the definitions of each process and what quantities change.

2. Recall that heat ($Q$) is non-zero for an isothermal process.

3. Compare each option to the properties of the processes listed.

Try solving on your own before revealing the answer!

Final Answer: Q for an isothermal process

Heat ($Q$) is non-zero for an isothermal process, since energy is transferred as heat while temperature remains constant.

Q7. Which process on the pV diagram below has a heat of $Q = 15$ L-atm? The gas is monatomic.

Background

Topic: Thermodynamics – Heat Transfer in Cyclic Processes

This question tests your ability to interpret a pV diagram and identify which process corresponds to a specific heat transfer.

Key Terms and Formulas:

• Heat ($Q$): Energy transferred to/from the system

• Monatomic gas: Use $C_v$ and $C_p$ as needed

• pV diagram: Visual representation of processes

Step-by-Step Guidance

1. Examine the pV diagram and identify the direction and type of each process (A to B, B to C, C to A).

2. Recall that heat transfer depends on the process (isothermal, isobaric, adiabatic, etc.).

3. Determine which process involves a heat transfer of 15 L-atm.

4. Set up the calculation for heat transfer for each process using the appropriate formula.

Try solving on your own before revealing the answer!

Final Answer: Process B to C

The process from B to C on the pV diagram corresponds to a heat transfer of 15 L-atm.

Q8. $n = 2$ mol of monatomic ideal gas expands slowly and adiabatically at an initial pressure of 5 atm from 10 L to 20 L. What is the final pressure of the gas?

Background

Topic: Adiabatic Expansion of Ideal Gases

This question tests your understanding of adiabatic processes and how pressure changes when volume changes for a monatomic ideal gas.

Key Terms and Formulas:

• Adiabatic process: $PV^{\gamma} = \text{constant}$

• For monatomic gas: $\gamma = \frac{5}{3}$

• $P_1, V_1$ = initial pressure and volume

• $P_2, V_2$ = final pressure and volume

Step-by-Step Guidance

1. Write down the initial and final values: $P_1 = 5$ atm, $V_1 = 10$ L, $V_2 = 20$ L.

2. Recall the adiabatic relation: $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.

3. Rearrange to solve for $P_2$: $P_2 = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma}$.

4. Set up the calculation for $P_2$ using $\gamma = \frac{5}{3}$.

Try solving on your own before revealing the answer!

Final Answer: 1.58 atm

The final pressure after adiabatic expansion is 1.58 atm. The pressure drops as the volume increases.

Q9. What is the heat for the diagonal line process illustrated on the pV diagram below? The gas is monatomic.

Background

Topic: Thermodynamics – Heat Transfer in pV Processes

This question tests your ability to calculate heat transfer for a process shown as a diagonal line on a pV diagram.

Key Terms and Formulas:

• First Law of Thermodynamics: $\Delta E_{th} = Q - W$

• Work ($W$): Area under the curve in pV diagram

• Monatomic gas: $C_v = \frac{3}{2}R$

Step-by-Step Guidance

1. Identify the initial and final states on the pV diagram.

2. Calculate the change in internal energy using $\Delta E_{th} = n C_v \Delta T$.

3. Calculate the work done using the area under the process curve.

4. Set up the equation for heat: $Q = \Delta E_{th} + W$.

Try solving on your own before revealing the answer!

Final Answer: 15 J

The heat transferred during the diagonal process is 15 J.

Q10. Suppose an ideal gas increases its temperature by 36°F. What is this temperature change in Celsius?

Background

Topic: Temperature Conversion

This question tests your ability to convert a temperature change from Fahrenheit to Celsius.

Key Terms and Formulas:

• Temperature change conversion: $\Delta T_C = \frac{5}{9} \Delta T_F$

• $\Delta T_F$ = change in Fahrenheit

• $\Delta T_C$ = change in Celsius

Step-by-Step Guidance

1. Write down the temperature change in Fahrenheit: $\Delta T_F = 36$°F.

2. Use the conversion formula: $\Delta T_C = \frac{5}{9} \Delta T_F$.

3. Set up the calculation for $\Delta T_C$.

Try solving on your own before revealing the answer!

Final Answer: 20°C

A temperature change of 36°F corresponds to a change of 20°C.

Q11. Two chambers of ideal gas A and B at the same temperature are separated by a pipe as illustrated in the image below. Chamber A has a volume $V_A = 4$ m3 with $n_A = 5$ mol of gas. Chamber B has a volume $V_B = 6$ m3 with $n_B = 3$ mol of ideal gas. What is a correct statement when the valve is opened?

Background

Topic: Gas Flow and Pressure Differences

This question tests your understanding of how gas flows between chambers based on pressure differences, given equal temperature.

Key Terms and Formulas:

• Ideal Gas Law: $P = \frac{nRT}{V}$

• Gas flows from high pressure to low pressure

Step-by-Step Guidance

1. Calculate the pressure in each chamber using $P = \frac{nRT}{V}$.

2. Compare the pressures to determine the direction of gas flow.

3. Recall that gas flows from higher to lower pressure.

Try solving on your own before revealing the answer!

Final Answer: Gas flows from A to B

Since chamber A has higher pressure, gas will flow from A to B when the valve is opened.

Q12. For the Stirling engine illustrated in the pV diagram below, what is the fuel heat per cycle, $Q_H$? The gas is monatomic.

Background

Topic: Heat Engines – Stirling Cycle

This question tests your understanding of the Stirling engine and how to calculate the heat input per cycle.

Key Terms and Formulas:

• Stirling cycle: Consists of isothermal and isochoric processes

• Heat input ($Q_H$): Occurs during the isothermal expansion

• For isothermal process: $Q = nRT \ln \left(\frac{V_f}{V_i}\right)$

Step-by-Step Guidance

1. Identify the isothermal expansion segment on the pV diagram.

2. Set up the formula for heat input during isothermal expansion.

3. Plug in the values for $n$, $R$, $T$, and the volume ratio.

Try solving on your own before revealing the answer!

Final Answer: $Q_H = nRT_i (1 + \ln 2)$

The heat input per cycle for the Stirling engine is $nRT_i (1 + \ln 2)$, where $T_i$ is the initial temperature.

Q13. A heat engine has a thermal efficiency of $e = 1/3$. If the waste heat per cycle is $Q_C = -20$ J, what is the work done by the engine per cycle, $W_{eng}$?

Background

Topic: Heat Engines – Efficiency and Work

This question tests your understanding of the relationship between efficiency, heat input, waste heat, and work output in a heat engine.

Key Terms and Formulas:

• Efficiency: $e = \frac{W_{eng}}{Q_H}$

• Waste heat: $Q_C$

• First law: $Q_H = W_{eng} + |Q_C|$

Step-by-Step Guidance

1. Write down the efficiency formula: $e = \frac{W_{eng}}{Q_H}$.

2. Express $Q_H$ in terms of $W_{eng}$ and $Q_C$.

3. Set up the equations and solve for $W_{eng}$.

Try solving on your own before revealing the answer!

Final Answer: 10 J

The work done by the engine per cycle is 10 J.

Q14. The closed loop process on the pV diagram illustrated below best represents what type of device?

Background

Topic: Thermodynamic Cycles and Devices

This question tests your ability to identify the type of device represented by a closed loop on a pV diagram.

Key Terms and Formulas:

• Closed loop: Indicates a cyclic process

• Heat engine: Converts heat to work

• Carnot engine: Idealized heat engine

• Heat pump: Moves heat from cold to hot reservoir

Step-by-Step Guidance

1. Examine the direction of the cycle on the pV diagram.

2. Recall the characteristics of heat engines, Carnot engines, and heat pumps.

3. Determine which device matches the process shown.

Try solving on your own before revealing the answer!

Final Answer: Heat engine

The closed loop process represents a heat engine, which converts heat into work in a cyclic process.

Q15. Suppose we replace the isochoric processes in the Stirling engine below with adiabatic processes so that we have a Carnot engine. What would be its Carnot efficiency?

Background

Topic: Carnot Engine Efficiency

This question tests your understanding of the Carnot efficiency formula and how it applies to idealized heat engines.

Key Terms and Formulas:

• Carnot efficiency: $e_{Carnot} = 1 - \frac{T_C}{T_H}$

• $T_H$ = temperature of hot reservoir

• $T_C$ = temperature of cold reservoir

Step-by-Step Guidance

1. Identify the temperatures of the hot and cold reservoirs from the pV diagram.

2. Set up the Carnot efficiency formula: $e_{Carnot} = 1 - \frac{T_C}{T_H}$.

3. Plug in the values for $T_C$ and $T_H$.

Try solving on your own before revealing the answer!

Final Answer: 1/2

The Carnot efficiency for this engine is 1/2, meaning it converts 50% of the heat input into work.