Solved Problems in Basic Physics

1. Ideal gases in a closed container initially have volume V and pressure P. If the final pressure is 4P and the volume is kept constant, what is the ratio of the initial kinetic energy with the final kinetic energy.

Known :

Initial pressure (P₁) = P

Final pressure (P₂) = 4P

Initial volume (V₁) = V

Final volume (V₂) = V

Wanted: The ratio of the initial kinetic energy with the final kinetic energy (KE₁ : KE₂)

Solution :

The relation between pressure (P), volume (V) and kinetic energy (KE) of ideal gases :

The ratio of the initial kinetic energy with the final kinetic energy :

2. What is the average translational kinetic energy of molecules in an ideal gas at 57^oC.

Known :

Temperature of gas (T) = 57^oC + 273 = 330 Kelvin

Boltzmann‘s constant (k) = 1.38 x 10^-23 Joule/Kelvin

Wanted: The average translational kinetic energy

Solution :

The relation between kinetic energy (KE) and the temperature of the gas (T) :

The average translational kinetic energy :

3. A gas at 27^oC in a closed container. If the kinetic energy of the gas increases 2 times the initial kinetic energy, thus the final temperature of the gas is…

Known :

Initial temperature (T₁) = 27^oC + 273 = 300 K

Initial kinetic energy = KE

Final kinetic energy = 4 KE

Wanted: The final temperature (T₂)

Solution :

4. An ideal gas is in a closed container, is heated so that the final average velocity of particles of gas increases by 3 times the initial average velocity. If the initial gas temperature is 27^oC, then the final temperature of the ideal gas is…

Known :

Initial temperature = 27^oC + 273 = 300 Kelvin

Initial velocity = v

Final velocity = 2v

Wanted : The final temperature of ideal gas

Solution :

The final average velocity = 2 x the initial average velocity

5. Three moles of gas are in a 36 liters volume space. Each gas molecule has a kinetic energy of 5 x 10^-21 Joule. Universal gas constant = 8.315 J/mole.K and Boltzmann’s constant = 1.38 x 10^-23 J/K. What is the gas pressure in the container.

Known :

Number of moles (n) = 3 moles

Volume = 36 liters = 36 dm³ = 36 x 10^-3 m³

Boltzmann’s constant (k) = 1.38 x 10^-23 J/K

Kinetic energy (KE) = 5 x 10^–21 Joule

Universal gas constant (R) = 8.315 J/mole.K

Wanted : Gas pressure (P)

Solution :

Calculate the temperature using the equation of kinetic energy of gas.

Calculate the gas pressure using th equation of ideal gas law (in number of moles, n) :

P V = n R T

P (36 x 10^-3) = (3)(8.315)(241.5)

P (36 x 10^-3) = 6024.22

The gas pressure is 1.67 x 10⁵ Pascal or 1.67 atmospheres.

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1. Ideal gases in a closed container initially have volume V and temperature T. The final temperature is 5/4T and the final pressure is 2P. What is the final volume of the gas?

Known :

Initial volume (V₁) = V

Initial temperature (T₁) = T

Final temperature (T₂) = 5/4 T

Initial pressure (P₁) = P

Final pressure (P₂) = 2P

Wanted: Final volume (V₂)

Solution :

2. Determine the volume of 2.00 moles of gases (ideal gas) at STP. STP = Standard Temperature and Pressure.

Known :

Moles of gas (n) = 2 moles

Standard temperature (T) = 0 ^oC = 0 + 273 = 273 Kelvin

Standard pressure (P) = 1 atm = 1.013 x 10⁵ Pa

Universal gas constant (R) = 8.315 Joule/mole.Kelvin

Wanted : Volume of gases (V)

Solution :

Equation of Ideal gas law (in the number of moles, n)

Volume 2 moles of gases is 44.8 liters.

Volume 1 mol of gases is 45.4 liters / 2 = 22.4 liters.

Volume 1 mol of any gases is 22.4 liters.

3. 4 liters of oxygen gas has a temperature of 27°C and pressure of 2 atm (1 atm = 10⁵ Pa) in a closed container. Universal gas constant (R) = 8.314 J.mole⁻¹.K⁻¹ and Avogadro’s number (N_A) = 6.02 x 10²³ molecules/mole. What are the molecules of oxygen gases in the container?

Known :

Volume of gases (V) = 4 liters = 4 dm³ = 4 x 10^-3 m³

Temperature of gases (T) = 27^oC = 27 + 273 = 300 Kelvin

Pressure of gases (P) = 2 atm = 2 x 10⁵ Pa

Universal gas constant (R) = 8.314 J.mole⁻¹.K⁻¹

Avogadro’s number (N_A) = 6.02 x 10²³

Wanted : What is the molecules of oxygen gases in the container (N)

Solution :

In 1 mole oxygen gases, there are 1.93 x 10²³ oxygen molecules.

4. A container containing a neon gas (Ne, atomic mass = 20 u) at standard temperature and pressure (STP) has a volume of 2 m³. Determine the mass of the neon gas!

Known :

Atomic mass of neon = 20 gram/mole = 0,02 kg/mole

Standard temperature (T) = 0^oC = 273 Kelvin

Standard pressure (P) = 1 atm = 1.013 x 10⁵ Pascal

Volume (V) = 2 m³

Wanted : mass (m) of neon gas

Solution :

At standard temperature and pressure (STP), 1 mole of any gases, include neon gas, have volume 22.4 liters = 22.4 dm³ = 0.0448 m³.

In the volume of 2 m³, there are 44.6 moles of neon gas.

Relative atomic mass of neon gas is 20 gram/mole.

This means that in 1 mole there are 20 grams or 0.02 kg of neon gas. Because in 1 mol there are 0.02 kg of neon gas then in 44.6 mole there are 44.6 moles x 0.02 kg/mole = 0.892 kg = 892 gram of neon gases.

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1. Ideal gases initially have pressure P and temperature T. The gas undergoes the isochoric process so that the final pressure becomes 4 times the initial pressure. What is the final temperature of the gas?

Known :

Initial pressure (P₁) = P

Final pressure (P₂) = 4P

Initial temperature (T₁) = T

Wanted: Final temperature (T₂)

Solution :

The formula of Gay-Lussac’s law :

The final temperature becomes 4 times the initial temperature.

2. In a closed container, ideal gases initially have a temperature of 27^oC. If the final pressure becomes 2 times the initial pressure, what is the final temperature?

Known :

Initial pressure (P₁) = P

Final pressure (P₂) = 2P

Initial temperature (T₁) = 27^oC + 273 = 300 K

Wanted: Final temperature (T₂)

Solution :

3. A tire is filled to a gauge pressure of 2 atm at 27°C. After a drive, the temperature within the tire rises to 47°C. What is the pressure within the tire now?

Known :

The atmospheric pressure = 1 atm = 1 x 10⁵ Pa

The initial gauge pressure = 2 atm = 2 x 10⁵ Pa

The initial absolute pressure (P₁) = 1 atm + 2 atm = 3 atm = 3 x 10⁵ Pa

The initial temperature (T₁) = 27^oC + 273 = 300 K

The final temperature (T₁) = 47^oC + 273 = 320 K

Wanted : The final gauge temperature

Solution :

The final gauge pressure = final absolute pressure – atmospheric pressure

The final gauge pressure = 3.2 atm – 1 atm

The final gauge pressure = 2.2 atm

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1. In a closed container, the gas expands so that the final volume becomes 3 times the initial volume (V = initial volume, T = initial temperature). What is the final temperature?

Known :

Initial volume (V₁) = V

Final volume (V₂) = 3V

Initial temperature (T₁) = T

Wanted: Final temperature (T₂)

Solution :

The formula of Charles’s law :

The final temperature of gases becomes 3 times the initial temperature.

2. Ideal gases initially have volume V and temperature T. If the gas undergoes the isobaric process so that the temperature becomes 2 times the initial temperature then the final volume of gases is…

Known :

Initial volume (V₁) = V

Initial temperature (T₁) = T

Final temperature (T₂) = 2T

Wanted: final volume (V₂)

Solution :

The final volume of gases becomes 2 times the initial volume.

3. In a closed container, ideal gases initially have a volume of 2 liters and temperature of 27^oC. If the final volume of gases becomes 3 liters then the final temperature is…

Known :

Initial volume (V₁) = 2 liters = 2 dm³ = 2 x 10^-3 m³

Final volume (V₂) = 3 liters = 3 dm³ = 3 x 10^-3 m³

Initial temperature (T₁) = 27^oC + 273 = 300 K

Wanted : Final temperature (T₂)

Solution :

The final temperature is 177^oC or 177 + 273 = 450 Kelvin.

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1. Some ideal gases initially have pressure P and volume V. If the gas undergoes isothermal process so that the final pressure becomes 4 times the initial pressure, then the final volume of gas is…

Known :

Initial pressure (P₁) = P

Final pressure (P₂) = 4P

Initial volume (V₁) = V

Wanted: Final volume (V₂)

Solution :

The formula of Boyle’s law :

P V = constant

P₁ V₁ = P₂ V₂

(P)(V) = (4P)(V₂)

V = 4 V₂

V₂ = V / 4 = ¼ V

The final volume of gases is ¼ times the initial volume.

2. In a closed container, the gas expands so that the final volume becomes 2 times the initial volume (V = initial volume, P = initial pressure). The final pressure of gases is…

Known :

Initial pressure (P₁) = P

Initial volume (V₁) = V

Final volume (V₂) = 2V

Wanted : Final pressure (P₂)

Solution :

P₁ V₁ = P₂ V₂

P V = P₂ (2V)

P = P₂ (2)

P₂ = P / 2 = ½ P

The gases pressure becomes ½ times the initial pressure.

3. In a closed container, gases having a pressure of 2 atm and a volume of 1 liter. If gas pressure becomes 4 atm then gas volume becomes …

Known :

Initial pressure (P₁) = 2 atm = 2 x 10⁵ Pa

Final pressure (P₂) = 4 atm = 4 x 10⁵ Pa

Initial volume (V₁) = 1 liter = 1 dm³ = 1 x 10^-3 m³

Wanted : Final volume (V₂)

Solution :

P₁ V₁ = P₂ V₂

(2 x 10⁵)(1 x 10^-3) = (4 x 10⁵) V₂

(1)(1 x 10^-3) = (2) V₂

1 x 10^-3 = (2) V₂

V₂ = ½ x 10^-3

V₂ = 0.5 x 10^-3 m³ = 0.5 dm³ = 0.5 liters

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1. Three capacitors, C₁ = 2 μF, C₂ = 4 μF, C₃ = 4 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination.

Known :

Capacitor C₁ = 2 μF

Capacitor C₂ = 4 μF

Capacitor C₃ = 4 μF

Wanted : The equivalent capacitance (C)

Solution :

Capacitor C₂ and C₃ connected in parallel. The equivalent capacitance :

C_P = C₂ + C₃ = 4 + 4 = 8 μF

Capacitor C₁ and C_p connected in series. The equivalent capacitance :

1/C = 1/C₁ + 1/C_P = 1/2 + 1/8 = 4/8 + 1/8 = 5/8

C = 8/5 μF

2. Five capacitors, C₁ = 2 μF, C₂ = 4 μF, C₃ = 6 μF, C₄ = 5 μF, C₅ = 10 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination.

Known :

Capacitor C₁ = 2 μF

Capacitor C₂ = 4 μF

Capacitor C₃ = 6 μF

Capacitor C₄ = 5 μF

Capacitor C₅ = 10 μF

Wanted : The equivalent capacitance (C)

Solution :

Capacitor C₂ and C₃ are connected in parallel. The equivalent capacitance :

C_P = C₂ + C₃

C_P = 4 + 6

C_P = 10 μF

Capacitor C₁, C_P, C₄ and C₅ are connected in series. The equivalent capacitance :

1/C = 1/C₁ + 1/C_P + 1/C₄ + 1/C₅

1/C = 1/2 + 1/10 + 1/5 + 1/10

1/C = 5/10 + 1/10 + 2/10 + 1/10

1/C = 9/10

C = 10/9 μF

3. C₁ = 3 μF, C₂ = 4 μF and C₃ = 3 μF, are connected in series and parallel. Determine the electric energy on the circuits.

Known :

Capacitor C₁ = 3 μF

Capacitor C₂ = 4 μF

Capacitor C₃ = 3 μF

Wanted : The equivalent capacitance (C)

Solution :

Capacitor C₂ and C₃ are connected in parallel. The equivalent capacitance :

C_P = C₂ + C₃

C_P = 4 + 3

C_P = 7 μF

Capacitor C₁ and C_P are connected in series. The equivalent capacitance :

1/C = 1/C₁ + 1/C_P

1/C = 1/3 + 1/7

1/C = 7/21 + 3/21

1/C = 10/21

C = 21/10

C = 2.1 μF

C = 2.1 x 10^-6 F

The electric energy on the circuits :

E = ½ C V²

E = ½ (2.1 x 10^-6)(12²)

E = ½ (2.1 x 10^-6)(144)

E = (2.1 x 10^-6)(72)

E = 151.2 x 10^-6 Joule

E = 1.5 x 10^-4 Joule

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1. Four capacitors, C₁ = 2 μF, C₂ = 1 μF, C₃ = 3 μF, C₄ = 4 μF, are connected in series. Determine the capacitance of a single capacitor that will have the same effect as the combination.

Known :

Capacitor C₁ = 2 μF

Capacitor C₂ = 1 μF

Capacitor C₃ = 3 μF

Capacitor C₃ = 4 μF

Wanted : The equivalent capacitance

Solution :

The equivalent capacitance :

1/C = 1/C₁ + 1/C₂ + 1/C₃ + 1/C₄

1/C = 1/2 + 1/1 + 1/3 + 1/4

1/C = 6/12 + 12/12 + 4/12 + 3/12

1/C = 25/12

C = 12/25

C = 0.48

The equivalent capacitance of the entire combination is 0.48 μF.

2. Determine the charge on capacitor C₁ if the potential difference between P and Q is 12 Volt…

Known :

Capacitor C₁ = 10 μF = 10 x 10^-6 F

Capacitor C₂ = 20 μF = 20 x 10^-6 F

Potential difference (V) = 12 Volt

Wanted : the charge on capacitor C₁ (Q₁)

Solution :

The equivalent capacitance :

1/C = 1/C₁ + 1/C₂

1/C = 1/10 + 1/20 = 2/20 + 1/20 = 3/20

C = 20/3 μF = (20/3) x 10^-6 F

Electric charge on the equivalent capacitor :

Q = (C)(V) = (20/3)(12)(10^-6) = 80 x 10^-6 C

Q = 80 μC

Capacitors are connected in series so that electric charge on the equivalent capacitors = electric charge on capacitor C₁ = electric charge on capacitor C₂.

The electric charge on capacitor C₁ is 80 μC.

3. Two capacitors, C₁ = 2 μF and C₂ = 4 μF, are connected in series. The capacitors are charged. The potential difference on capacitor C₁ is 2 Volt. The electric charge on capacitor C₂ is…

Known :

Capacitor C₁ = 2 μF = 2 x 10^-6 F

Capacitor C₂ = 4 μF = 4 x 10^-6 F

The potential difference on capacitor C₁ (V₁) = 2 Volt

Wanted : Electric charge on capacitor C₂.

Solution :

Electric charge on capacitor C₁ :

Q₁ = C₁ V₁ = (2 x 10^-6)(2) = 4 x 10^-6 C

Q₁ = 4 μC

Capacitors are connected series so that electric charge on capacitor C₁ = electric charge on capacitor C₂.

The charge on capacitor C₂ is 4 μC.

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1. Four capacitors, C₁ = 2 μF, C₂ = 1 μF, C₃ = 3 μF, C₄ = 4 μF, are connected in parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination.

Known :

Capacitor C₁ = 2 μF

Capacitor C₂ = 1 μF

Capacitor C₃ = 3 μF

Capacitor C₃ = 4 μF

Wanted : The equivalent capacitance

Solution :

The equivalent capacitance :

C = C₁ + C₂ + C₃

C = 4 μF + 2 μF + 3 μF = 9 μF

The equivalent capacitance of the entire combination is 9 μF.

2. Determine the charge on capacitor C₂ if the potential difference between point A and B is 9 Volt…

Known :

Capacitor C₁ = 20 μF = 20 x 10^-6 F

Capacitor C₂ = 30 μF = 30 x 10^-6 F

Potential difference between point A and B (V_AB) = 9 Volt

Wanted : the charge on capacitor C₂ (Q₂)

Solution :

Potential difference :

Capacitors are connected in parallel so that the potential difference between A and B (V_AB) = the potential difference on capacitor C₁ (V₁) = the potential difference on capacitor C₂ (V₂) = 9 Volt.

Electric charge on capacitor C₂ :

Q₂ = C₂ V₂ = (30 x 10^-6)(9) = 270 x 10^-6 C

Q₂ = 270 μC

The electric charge on capacitor C₂ is 270 μC.

3. Three capacitors, C₁ = 4 μF, C₂ = 2 μF, C₃ = 3 μF, are connected in parallel. The capacitor are charged. The potential difference on capacitor C₂ is 4 Volt. Determine

(a) Electric charge on capacitor C₁, C₂ and C₃

(b) Electric charge on the equivalent capacitor of the entire combination

Known :

Capacitor C₁ = 4 μF = 4 x 10^-6 F

Capacitor C₂ = 2 μF = 2 x 10^-6 F

Capacitor C₃ = 3 μF = 3 x 10^-6 F

Potential difference on capacitor C₂ (V₂) = 4 Volt

Wanted : Electric charge on capacitor C₃ (Q₃)

Solution :

(a) Electric charge on capacitor C₃

The potential difference on capacitor C₃ :

Capacitors are connected in parallel so that the potential difference on capacitor C₃ (V₃) = the potential difference on capacitor C₂ (V₂) = the potential difference on capacitor C₁ (V₁) = the potential difference on equivalent capacitor (V) = 4 Volt

Electric charge on capacitor C₁ :

Q₁ = C₁ V₁ = (4 x 10^-6)(4) = 16 x 10^-6 C

Q₁ = 16 μC

Electric charge on capacitor C₂ :

Q₂ = C₂ V₂ = (2 x 10^-6)(4) = 8 x 10^-6 C

Q₂ = 8 μC

Electric charge on capacitor C₃ :

Q₃ = C₃ V₃ = (3 x 10^-6)(4) = 12 x 10^-6 C

Q₃ = 12 μC

(b) Electric charge on equivalent capacitor

Q = Q₁ + Q₂ + Q₃

Q = 16 μC + 8 μC + 12 μC = 36 μC

Alternative solution :

The equivalent capacitance :

C = C₁ + C₂ + C₃

C = 4 μF + 2 μF + 3 μF = 9 μF

C = 9 x 10^-6 F

The potential difference on the equivalent capacitor :

V₁ = V₂ = V₃ = V = 4 Volt

The electric charge on the equivalent capacitor :

Q = C V = (9 x 10^-6)(4) = 36 x 10^-6 C

Q = 36 μC

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