Question 1

A sample of an ideal gas is compressed by a piston from 10 m³ to 5 m³ and simultaneously cooled from 273 $\degree$C to 0 $\degree$C. As a result there is:

Accepted Answer

E

Question 2

An automobile tire is pumped up to a gauge pressure of 2.0 * 10⁵ Pa when the temperature is 27$\degree$C. What is its gauge pressure after the car has been running on a hot day so that the tire temperature is 77$\degree$C? Assume that the volume remains fixed and take atmospheric pressure to be 1.013 *10⁵ Pa.

Accepted Answer

A

Question 3

Air enters a hot-air furnace at 7 $\degree$C and leaves at 77 $\degree$C. If the pressure does not change each entering cubic foot of air expands to:

Accepted Answer

Using Charles's Law (V1/T1 = V2/T2), the volume of air expands proportionally with temperature when pressure is constant. Converting temperatures to Kelvin and solving gives approximately 1.25 m³.

Question 4

A real gas is changed slowly from state 1 to state 2. During this process no work is done on or by the gas. This process must be:

Accepted Answer

If no work is done on or by the gas, the volume must remain constant, indicating an isovolumic (isochoric) process.

Question 5

A 2-m³ weather balloon is loosely filled with helium at 1 atm (76 cm Hg) and at 27 $\degree$C. At an elevation of 20,000 ft, the atmospheric pressure is down to 38 cm Hg and the helium has expanded, being under no constraint from the confining bag. If the temperature at this elevation is -48 $\degree$C, the gas volume (in m³) is:

Accepted Answer

The answer of A 2-m³ weather balloon is loosely filled...

Question 6

An ideal gas occupies 12 liters at 293 K and 1 atm (76 cm Hg). Its temperature is now raised to 373 K and its pressure increased to 215 cm Hg. The new volume is:

Accepted Answer

Using the ideal gas law and the combined gas law, the new volume can be calculated as follows: V2 = V1 * (P1/P2) * (T2/T1) = 12 * (76/215) * (373/293) = 5.4 liters.

Question 7

Two identical rooms in a house are connected by an open doorway. The temperatures in the two rooms are maintained at different values. Which room contains more air?

Accepted Answer

The amount of air in a room is determined by its volume, and the size of the room is the same in both cases. Temperature is directly proportional to the volume. When the temperature is increased, the volume or size of the room increases. As a result, the room with a lower temperature has more air.

Question 8

An air bubble doubles in volume as it rises from the bottom of a lake (1000 kg/m³). Ignoring any temperature changes, the depth of the lake is:

Accepted Answer

The answer of An air bubble doubles in volume as...

Question 9

Over 1 cycle of a cyclic process in which a system does net work on its environment:

Accepted Answer

In a cyclic process, the system returns to its initial state at the end of the cycle. This means that all state functions, including pressure, volume, temperature, and internal energy, return to their initial values. Therefore, the changes in these properties over one complete cycle are zero.

Question 10

273 cm³ of an ideal gas is at 0 $\degree$C. It is heated at constant pressure to 10 $\degree$C. It will now occupy:

Accepted Answer

According to Charles's Law, the volume of a gas is directly proportional to its temperature at constant pressure. Therefore, V1/T1 = V2/T2. Converting the temperatures to Kelvin: T1 = 273K T2 = 283K Substituting in the formula: V1/273 = V2/283 V2 = (V1/273)*283 Since pressure is held constant, we can assume that the initial volume is equal to the final volume: V1 = V2 V1 = (273/283)*V2 V2 = (283/273)*V1 V2 = 1.037*V1 Therefore, the final volume is 1.037 times the initial volume. V2 = 1.037*273 cm3 = 283 cm3 Therefore, the answer is C, 283 cm³.

Question 11

An isothermal process for an ideal gas is represented on a p-V diagram by:

Accepted Answer

An isothermal process for an ideal gas is represented by a hyperbola on a p-V diagram, according to the ideal gas law PV = nRT, where P is pressure, V is volume, n is the amount of substance, R is the ideal gas constant, and T is temperature. In an isothermal process, the temperature (T) remains constant, so the product of pressure (P) and volume (V) remains constant, resulting in a hyperbolic curve.

Question 12

A given mass of gas is enclosed in a suitable container so that it may be maintained at constant volume. Under these conditions, there can be no change is what property of the gas?

Accepted Answer

Density remains constant because the volume and mass of the gas are constant.

Question 13

Oxygen (molar mass = 32 g) occupies a volume of 12 liters when its temperature is 20$\degree$C and its pressure is 1 atm. Using R = 0.082 liter . atm/mole. K, calculate the mass of the oxygen:

Accepted Answer

Using the ideal gas law, PV = nRT, we can calculate the number of moles of oxygen present:

n = (PV) / (RT)
n = (1 atm * 12 L) / (0.082 (L*atm/mol*K) * (20 + 273.15 K))
n = 0.484 moles

To calculate the mass of oxygen, we can use the molar mass:

mass = n * molar mass
mass = 0.484 moles * 32 g/mol
mass = 15.488 g

Therefore, the closest answer choice is C) 16 g.

Question 14

A real gas undergoes a process which can be represented as a curve on a p-V diagram. The work done by the gas during this process is:

Accepted Answer

The work done by a gas during a process represented on a p-V diagram is given by the area under the curve on the diagram, which is mathematically represented by the integral of pressure with respect to volume, or $\int p dV$. This corresponds to choice D.

Question 15

A quantity of an ideal gas is compressed to half its initial volume. The process may be adiabatic, isothermal or isobaric. Rank those three processes in order of the work required of an external agent, least to greatest.

Accepted Answer

The answer of A quantity of an ideal gas is...

Question 16

Use R = 8.2 *10^-⁵ m³ . atm/mol . K and N_A = 6.02 * 10²³ mol^-¹. The approximate number of air molecules in a 1 m³ volume at room temperature (300 K) and atomospheric pressure is:

Accepted Answer

We can use the ideal gas law to solve the problem: PV = nRT where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for n: n = PV/RT Substituting in the given values: - P = 1 atm = 1.01325 × 10^5 Pa - V = 1 m^3 - R = 8.2 × 10^-2 atm/mol K - T = 300 K We can also use the value given for N_A to convert from moles to number of molecules: N_A = 6.02 × 10^23 molecules/mol Putting it all together: n = (1.01325 × 10^5 Pa)(1 m^3)/(8.2 × 10^-2 atm/mol K)(300 K) = 4.978 × 10^25 molecules Number of molecules = n × N_A = (4.978 × 10^25 molecules)(6.02 × 10^23 molecules/mol) ≈ 2.999 × 10^49 Rounding to the nearest power of 10, we get approximately 2.5 × 10^25, which is answer choice C.

Question 17

The pressures p and volumes V of the five ideal gases,with the same number of molecules, are given below. Which has the highest theperatures?

Accepted Answer

The answer of The pressures p and volumes V of...

Question 18

An ideal gas undergoes an isothermal process starting with a pressure of 2*10⁵ Pa and a volume of 6 cm³. Which of the following might be the pressure and volume of the final state?

Accepted Answer

In an isothermal process, the temperature remains constant. Therefore, we can use the ideal gas law to find the final pressure and volume:

P1V1 = P2V2

where P1 = 2 x 10^5 Pa, V1 = 6 cm^3, and P2 and V2 are the final pressure and volume, respectively. Since the process is isothermal, we can also use the equation PV = nRT, where n is the number of moles of gas, R is the gas constant, and T is the temperature.

Setting the initial and final states equal, we have:

P1V1 = nRT

and

P2V2 = nRT

Dividing the second equation by the first, we get:

P2/P1 = V1/V2

Substituting the values we have:

P2/2 x 10^5 Pa = 6 cm^3/V2

P2V2 = 12 x 10^5 cm^3 Pa

Now, we can use the ideal gas law to find the final volume:

PV = nRT

Substituting P2 = 6 x 10^5 Pa, V2 = ?, and the other values:

6 x 10^5 Pa x V2 = n x R x T

Since the process is isothermal, T remains constant. Therefore, we can write:

6 x 10^5 Pa x V2 = n x R x T1

where T1 is the initial temperature.

We can simplify this equation using the fact that for an ideal gas, PV = nRT:

6 x 10^5 Pa x V2 = P1V1

Substituting the values we have:

6 x 10^5 Pa x V2 = 2 x 10^5 Pa x 6 cm^3

V2 = 2 cm^3

Therefore, the final state is:

Pressure = 6 x 10^5 Pa
Volume = 2 cm^3

The answer is choice D.

Question 19

It is known that 28 grams of a certain ideal gas occupy 22.4 liters at standard conditions (0 $\degree$C, 1 atm). The volume occupied by 42 grams of this gas at standard conditions is:

Accepted Answer

The given gas is an ideal gas, which means that its mass is directly proportional to its volume at constant temperature and pressure. Using this relationship, we can set up a proportion:

28 g / 22.4 L = 42 g / V

Solving for V, we get:

V = (42 g x 22.4 L) / 28 g = 33.6 L

Therefore, the volume occupied by 42 grams of the gas at standard conditions is 33.6 liters, choice C.

Question 20

In order that a single process be both isothermal and isobaric:

Accepted Answer

A change of phase, such as from a liquid to a gas or a solid to a liquid, can allow for a process to be both isothermal (constant temperature) and isobaric (constant pressure). This is because the energy added or lost during the phase change goes solely into breaking or forming intermolecular bonds, rather than changing the temperature or pressure of the system.

Quiz 19: The Kinetic Theory of Gases

A sample of an ideal gas is compressed by a piston from 10 m³ to 5 m³ and simultaneously cooled from 273 $\degree$ C to 0 $\degree$ C. As a result there is:

Air enters a hot-air furnace at 7 $\degree$ C and leaves at 77 $\degree$ C. If the pressure does not change each entering cubic foot of air expands to:

A real gas is changed slowly from state 1 to state 2. During this process no work is done on or by the gas. This process must be:

An ideal gas occupies 12 liters at 293 K and 1 atm (76 cm Hg) . Its temperature is now raised to 373 K and its pressure increased to 215 cm Hg. The new volume is:

Two identical rooms in a house are connected by an open doorway. The temperatures in the two rooms are maintained at different values. Which room contains more air?

An air bubble doubles in volume as it rises from the bottom of a lake (1000 kg/m³) . Ignoring any temperature changes, the depth of the lake is:

Over 1 cycle of a cyclic process in which a system does net work on its environment:

273 cm³ of an ideal gas is at 0 $\degree$ C. It is heated at constant pressure to 10 $\degree$ C. It will now occupy:

An isothermal process for an ideal gas is represented on a p-V diagram by:

A given mass of gas is enclosed in a suitable container so that it may be maintained at constant volume. Under these conditions, there can be no change is what property of the gas?

Oxygen (molar mass = 32 g) occupies a volume of 12 liters when its temperature is 20 $\degree$ C and its pressure is 1 atm. Using R = 0.082 liter . atm/mole. K, calculate the mass of the oxygen:

A real gas undergoes a process which can be represented as a curve on a p-V diagram. The work done by the gas during this process is:

A quantity of an ideal gas is compressed to half its initial volume. The process may be adiabatic, isothermal or isobaric. Rank those three processes in order of the work required of an external agent, least to greatest.

Use R = 8.2 10^-⁵ m³ . atm/mol . K and N_A = 6.02 10²³ mol^-¹. The approximate number of air molecules in a 1 m³ volume at room temperature (300 K) and atomospheric pressure is:

The pressures p and volumes V of the five ideal gases,with the same number of molecules, are given below. Which has the highest theperatures?

An ideal gas undergoes an isothermal process starting with a pressure of 2*10⁵ Pa and a volume of 6 cm³. Which of the following might be the pressure and volume of the final state?

It is known that 28 grams of a certain ideal gas occupy 22.4 liters at standard conditions (0 $\degree$ C, 1 atm) . The volume occupied by 42 grams of this gas at standard conditions is:

In order that a single process be both isothermal and isobaric:

Quiz 19: The Kinetic Theory of Gases

A sample of an ideal gas is compressed by a piston from 10 m3 to 5 m3 and simultaneously cooled from 273 °\degree° C to 0 °\degree° C. As a result there is:

Air enters a hot-air furnace at 7 °\degree° C and leaves at 77 °\degree° C. If the pressure does not change each entering cubic foot of air expands to:

A real gas is changed slowly from state 1 to state 2. During this process no work is done on or by the gas. This process must be:

An ideal gas occupies 12 liters at 293 K and 1 atm (76 cm Hg) . Its temperature is now raised to 373 K and its pressure increased to 215 cm Hg. The new volume is:

Two identical rooms in a house are connected by an open doorway. The temperatures in the two rooms are maintained at different values. Which room contains more air?

An air bubble doubles in volume as it rises from the bottom of a lake (1000 kg/m3) . Ignoring any temperature changes, the depth of the lake is:

Over 1 cycle of a cyclic process in which a system does net work on its environment:

273 cm3 of an ideal gas is at 0 °\degree° C. It is heated at constant pressure to 10 °\degree° C. It will now occupy:

An isothermal process for an ideal gas is represented on a p-V diagram by:

A given mass of gas is enclosed in a suitable container so that it may be maintained at constant volume. Under these conditions, there can be no change is what property of the gas?

Oxygen (molar mass = 32 g) occupies a volume of 12 liters when its temperature is 20 °\degree° C and its pressure is 1 atm. Using R = 0.082 liter . atm/mole. K, calculate the mass of the oxygen:

A real gas undergoes a process which can be represented as a curve on a p-V diagram. The work done by the gas during this process is:

A quantity of an ideal gas is compressed to half its initial volume. The process may be adiabatic, isothermal or isobaric. Rank those three processes in order of the work required of an external agent, least to greatest.

Use R = 8.2 *10-5 m3 . atm/mol . K and NA = 6.02 * 1023 mol-1. The approximate number of air molecules in a 1 m3 volume at room temperature (300 K) and atomospheric pressure is:

The pressures p and volumes V of the five ideal gases,with the same number of molecules, are given below. Which has the highest theperatures?

An ideal gas undergoes an isothermal process starting with a pressure of 2*105 Pa and a volume of 6 cm3. Which of the following might be the pressure and volume of the final state?

It is known that 28 grams of a certain ideal gas occupy 22.4 liters at standard conditions (0 °\degree° C, 1 atm) . The volume occupied by 42 grams of this gas at standard conditions is:

In order that a single process be both isothermal and isobaric:

A sample of an ideal gas is compressed by a piston from 10 m³ to 5 m³ and simultaneously cooled from 273 $\degree$ C to 0 $\degree$ C. As a result there is:

Air enters a hot-air furnace at 7 $\degree$ C and leaves at 77 $\degree$ C. If the pressure does not change each entering cubic foot of air expands to:

An air bubble doubles in volume as it rises from the bottom of a lake (1000 kg/m³) . Ignoring any temperature changes, the depth of the lake is:

273 cm³ of an ideal gas is at 0 $\degree$ C. It is heated at constant pressure to 10 $\degree$ C. It will now occupy:

Oxygen (molar mass = 32 g) occupies a volume of 12 liters when its temperature is 20 $\degree$ C and its pressure is 1 atm. Using R = 0.082 liter . atm/mole. K, calculate the mass of the oxygen:

Use R = 8.2 10^-⁵ m³ . atm/mol . K and N_A = 6.02 10²³ mol^-¹. The approximate number of air molecules in a 1 m³ volume at room temperature (300 K) and atomospheric pressure is:

An ideal gas undergoes an isothermal process starting with a pressure of 2*10⁵ Pa and a volume of 6 cm³. Which of the following might be the pressure and volume of the final state?

It is known that 28 grams of a certain ideal gas occupy 22.4 liters at standard conditions (0 $\degree$ C, 1 atm) . The volume occupied by 42 grams of this gas at standard conditions is: