The work done in a constant pressure expansion can be determined by:

The work done in a constant pressure expansion can be determined by:

A. P(V2 - V1)

B. P(V1 - V2)

C. (P1V1 - P2V2)/r - 1

D. PV x 2.3 log V1/V2

E. P1V1 - P2V2

Answer: A

If 10.4 m³ of air is compressed at a constant temperature to 5.1 m³ and 750 kPa gauge. Calculate how much work is done on the gas, assuming atmospheric pressure to be 100 kPa.

If 10.4 m³ of air is compressed at a constant temperature to 5.1 m³ and 750 kPa gauge. Calculate how much work is done on the gas, assuming atmospheric pressure to be 100 kPa.

A. 2727 J

B. 2850 J

C. 3051 J

D. 3089 J

E. 3128 J

Answer: D

Boyle's Law is one of constant:

Boyle's Law is one of constant:

A. Mass

B. Volume

C. Pressure

D. Temperature

E. Flow


Answer: D

The pressure-volume diagram of a typical air compressor will have a curve drawn:

The pressure-volume diagram of a typical air compressor will have a curve drawn:

A. As an adiabatic compression

B. As in an isothermal compression

C. Approximately half way between the adiabatic and isothermal curves

D. Almost as a straight line

E. From left to right in increasing value curved up


Answer: C

If the volume of a confined gas is constant then the absolute pressure is directly proportional to the absolute temperature according to:

If the volume of a confined gas is constant then the absolute pressure is directly proportional to the absolute temperature according to:

A. Boyle's Law

B. Bernoulli's' Law

C. Charles' Law

D. The General Gas Law

E. Gay-Lussac's Law


Answer: E

Find the characteristic constant for a gas if 112 kg of the gas has a volume of 5.7 m³ at 1500 kPa absolute and 27.5°C.

Find the characteristic constant for a gas if 112 kg of the gas has a volume of 5.7 m³ at 1500 kPa absolute and 27.5°C.

A. 0.254 kJ/kg K

B. 0.261 kJ/kg K

C. 0.271 kJ/kg K

D. 0.274 kJ/kg K

E. 0.291 kJ/kg K

Answer: A

If 24,500 kJ of work is done in a cylinder when 115 kg of a perfect gas is expanded isothermally at 210°C from 0.035 m³ to 6.9 m³. Find the characteristic constant for this gas.

If 24,500 kJ of work is done in a cylinder when 115 kg of a perfect gas is expanded isothermally at 210°C from 0.035 m³ to 6.9 m³. Find the characteristic constant for this gas.

A. 0.083 kJ/kg K

B. 0.198 kJ/kg K

C. 0.421 kJ/kg K

D. 0.441 kJ/kg K

E. 0.489 kJ/kg K


Answer: A

A gas at 1000 kPa gauge pressure and 30°C is transferred from a cylindrical vessel 1.5 m in diameter and 3 m long to another cylindrical vessel 2.5 m in diameter and 5 m long. If the new gauge pressure is 150 kPa, calculate the new temperature. Note: Assume atmospheric pressure to be 100 kPa for this calculation.

A gas at 1000 kPa gauge pressure and 30°C is transferred from a cylindrical vessel 1.5 m in diameter and 3 m long to another cylindrical vessel 2.5 m in diameter and 5 m long. If the new gauge pressure is 150 kPa, calculate the new temperature. Note: Assume atmospheric pressure to be 100 kPa for this calculation.

A. 350°C

B. 318°C

C. 45.9°C

D. 63.7°C

E. 436°C


Answer: C

If 0.25 m³ of a gas at 4000 kPa gauge pressure is expanded until the gauge pressure is 500 kPa and the expansion is polytropic with n = 1.35. Find the final volume the gas will occupy. Note: Assume atmospheric pressure = 100 kPa

If 0.25 m³ of a gas at 4000 kPa gauge pressure is expanded until the gauge pressure is 500 kPa and the expansion is polytropic with n = 1.35. Find the final volume the gas will occupy. Note: Assume atmospheric pressure = 100 kPa

A. 1.04 m³

B. 1.24 m³

C. 1.30 m³

D. 1.33 m³

E. 1.47 m³


Answer: A

The equation (PV/T) = mR or PV = mRT is called the:

The equation (PV/T) = mR or PV = mRT is called the:

A. Characteristic equation of a perfect gas

B. Characteristic constant of a perfect gas

C. Perfect gas law equation

D. Volume constant for perfect gases

E. Boyle's law


Answer: A

The Characteristic Constant of each perfect gas is

The Characteristic Constant of each perfect gas is

A. The same for all perfect gases

B. The same for all gases if their temperatures are the same

C. The same for all gases if their pressures are the same

D. Unique to that particular gas

E. Varying directly with temperature changes


Answer: D

The pressure-volume diagram of a typical air compressor will have a curve drawn:

The pressure-volume diagram of a typical air compressor will have a curve drawn:

A. As an adiabatic compression

B. As in an isothermal compression

C. Approximately half way between the adiabatic and isothermal curves

D. Almost as a straight line

E. From left to right in increasing value curved up

Answer: C

If 10.4 m³ of air is compressed at a constant temperature to 5.1 m³ and 750 kPa gauge. Calculate how much work is done on the gas, assuming atmospheric pressure to be 100 kPa.

If 10.4 m³ of air is compressed at a constant temperature to 5.1 m³ and 750 kPa gauge. Calculate how much work is done on the gas, assuming atmospheric pressure to be 100 kPa.

A. 2727 J

B. 2850 J

C. 3051 J

D. 3089 J

E. 3128 J


Answer: D

Compression and expansion in real life cases will always follow the polytropic process that is described by the formula(s):

Compression and expansion in real life cases will always follow the polytropic process that is described by the formula(s): 

1. P1V1 = P2V2
2. P1V1n = P2V2n
3. PVn = Constant
4. PV = Constant


A. 1, 2

B. 2, 3

C. 3, 4

D. 1, 3

E. 2, 4


Answer: B

The formula Work = (P1V1 - P2V2) / n - 1 is used to describe expansion during:

The formula Work = (P1V1 - P2V2) / n - 1 is used to describe expansion during:

A. An isothermal process

B. An isentropic process

C. A polytropic process

D. An adiabatic process

E. An enthalpy process


Answer: C

To "expand a gas" means the gas is:

To "expand a gas" means the gas is:

A. Always at constant volume

B. Always at constant pressure

C. Always at constant temperature

D. Able to perform useful work

E. Required to have work preformed on it


Answer: D

The work done in a constant pressure expansion can be determined by:

The work done in a constant pressure expansion can be determined by:

A. P(V2 - V1)

B. P(V1 - V2)

C. (P1V1 - P2V2)/r - 1

D. PV x 2.3 log V1/V2

E. P1V1 - P2V2


Answer: A

A perfect gas is compressed under conditions of constant temperature to a volume of 30 m³. If the final pressure of the gas is 450 kPa gauge and the initial volume was 135 m³, what was the initial pressure? (Assume atmospheric pressure to be 101.3 kPa)

A perfect gas is compressed under conditions of constant temperature to a volume of 30 m³. If the final pressure of the gas is 450 kPa gauge and the initial volume was 135 m³, what was the initial pressure? (Assume atmospheric pressure to be 101.3 kPa)

A. 122.5 kPa gauge

B. 21.2 kPa gauge

C. 223.8 kPa gauge

D. 101.3 kPa gauge

E. 124.8 kPa gauge


Answer: B

Gases such as air, nitrogen and oxygen can be roughly defined as perfect gases because they:

Gases such as air, nitrogen and oxygen can be roughly defined as perfect gases because they:

A. Condense rapidly during condition changes

B. Can expand without being heated

C. Remain in gaseous form during condition changes

D. Cannot be superheated

E. None of the above


Answer: C

The General Gas Law can be written using the formula/s:

The General Gas Law can be written using the formula/s: 

1. PV/T = C (constant)
2. P1V1/T2 = P2V2/T1
3. P1V1/T1 = P2V2/T2
4. P1V2/T1 = P2V1/T2


A. 1, 2

B. 2, 3

C. 3, 4

D. 1, 3

E. 2, 4


Answer: D