Question 1

Evidence that a gas consists mostly of empty space is the fact that:
A)the density of a gas becomes much greater when it is liquefied
B)gases exert pressure on the walls of their containers
C)gases are transparent
D)heating a gas increases the molecular motion
E)nature abhors a vacuum

Accepted Answer

When a gas is liquefied, its particles are brought closer together, resulting in a much greater density. This indicates that gas particles occupy a significant amount of space, meaning that a large portion of the gas is empty.

Question 2

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:
A)3.3 liters
B)5.4 liters
C)27 liters
D)46 liters
E)none of these

Accepted Answer

We can use the combined gas law to solve for the new volume:

(P₁V₁/T₁) = (P₂V₂/T₂)

We can convert the initial pressure to atm:

1 atm = 76 cm Hg * (1 atm / 760 cm Hg) = 0.1 atm

Similarly, we can convert the final pressure to atm:

215 cm Hg * (1 atm / 760 cm Hg) = 0.283 atm

Now we can plug in the values:

(0.1 L) / (293 K) = (0.283 atm * V₂) / (373 K)

Solving for V₂:

V₂ = (0.1 L * 373 K * 0.283 atm) / (293 K * 1)

V₂ = 5.4 L

Therefore, the new volume is 5.4 liters.

Question 3

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 in what property of the gas?
A)Pressure
B)Density
C)Molecular kinetic energy
D)Internal energy
E)Temperature

Accepted Answer

Since the volume is constant, the density of the gas cannot change. Pressure, molecular kinetic energy, internal energy, and temperature may all change depending on external factors or changes in the gas itself.

Question 4

Over 1 cycle of a cyclic process in which a system does net work on its environment:
A)the change in the pressure of the system cannot be zero
B)the change in the volume of the system cannot be zero
C)the change in the temperature of the system cannot be zero
D)the change in the internal energy of the system cannot be zero
E)none of the above

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 5

An automobile tire is pumped up to a gauge pressure of 2.0 * 10⁵ Pa when the temperature is 27°C.What is its gauge pressure after the car has been running on a hot day so that the tire temperature is 77°C? Assume that the volume remains fixed and take atmospheric pressure to be 1.013 *10⁵ Pa. A)1.6 * 10⁵ Pa B)2.3 *10⁵ Pa C)2.5 * 10⁵ Pa D)3.6 * 10⁵ Pa E)8.6 * 10⁵ Pa

Accepted Answer

The gauge pressure is given by:$$P_g = P - P_0$$where:* $$P_g$$ is the gauge pressure* $$P$$ is the absolute pressure* $$P_0$$ is the atmospheric pressureThe absolute pressure is given by:$$P = P_0 + P_g$$The volume of the tire is fixed, so the absolute pressure is proportional to the temperature. Therefore, we have:$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$where:* $$P_1$$ is the absolute pressure at temperature $$T_1$$* $$P_2$$ is the absolute pressure at temperature $$T_2$$Substituting the given values, we get:$$\frac{P_1}{273 K} = \frac{P_2}{350 K}$$Solving for $$P_2$$, we get:$$P_2 = \frac{350 K}{273 K} P_1$$Substituting the given values for $$P_1$$ and $$P_0$$, we get:$$P_2 = \frac{350 K}{273 K} (2.0 imes 10^5 Pa + 1.013 imes 10^5 Pa)$$$$P_2 = 2.5 imes 10^5 Pa$$Therefore, the gauge pressure is:$$P_g = P_2 - P_0 = 2.5 imes 10^5 Pa - 1.013 imes 10^5 Pa = 1.487 imes 10^5 Pa$$The closest answer choice is C) 2.5 F1F1F1S1*F1F1F10 10⁵ Pa.

Question 6

Platinum has a molar mass of 195 g/mol.If you have a ring that contains 2.3 g of platinum, how many moles does it contain? A)0.012 mol B)85 mol C)450 mol D)7.2 x 10²¹ mol E)1.4 x 10²⁴ mol

Accepted Answer

To find the number of moles, divide the mass of the platinum by its molar mass: $ \frac{2.3 \, \text{g}}{195 \, \text{g/mol}} = 0.0118 \, \text{mol} $, which rounds to 0.012 mol.

Question 7

A 2.0-m³ weather balloon is loosely filled with helium at 1 atm (76 cm Hg)and at 27°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°C, the gas volume is: A)0.75 m³ B)1.3 m³ C)3.0 m³ D)4.0 m³ E)5.3 m³

Accepted Answer

To solve this problem, we can use the combined gas law, which is $P_1V_1/T_1 = P_2V_2/T_2$, where $P$ is pressure, $V$ is volume, and $T$ is temperature in Kelvin. Given that the initial conditions are 1 atm (76 cm Hg) and 20°C (which is 293K), and the final conditions are 38 cm Hg and -48°C (-48°C = 225K), we can solve for $V_2$. The initial volume $V_1$ is 2.0 m³. Converting pressures to the same units and temperatures to Kelvin, we can solve for $V_2$, which results in approximately 3.0 m³.

Question 8

Avogadro's number is: A)6.02 x 10¹⁸ mol^-1 B)6.02 x 10²⁰ mol^-1 C)6.02 x 10²² mol^-1 D)6.02 x 10²³ mol^-1 E)6.02 x 10²⁴ mol^-1

Accepted Answer

Avogadro's number, which is a fundamental constant in chemistry, is 6.02 x 10^23 mol^-1, representing the number of atoms, ions, or molecules in one mole of substance.

Question 9

It is known that 28 grams of a certain ideal gas occupy 22.4 liters at standard conditions (0°C, 1 atm).The volume occupied by 42 grams of this gas at standard conditions is:
A)14.9 liters
B)22.4 liters
C)33.6 liters
D)42 liters
E)more data are needed

Accepted Answer

The amount of gas (in moles) is directly proportional to the mass of the gas. Therefore, 42 grams of the gas is 1.5 times the amount of gas as 28 grams. As the volume is also directly proportional to the amount of gas (when pressure and temperature are constant), the volume of 42 grams of the gas at standard conditions would be 1.5 times the volume of 28 grams of the gas, which is 33.6 liters.

Question 10

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?
A)the room with higher temperature
B)the room with lower temperature
C)the room with higher pressure
D)neither because both have the same pressure
E)neither because both have the same volume

Accepted Answer

The answer of Two identical rooms in a house are...

Question 11

Oxygen (molar mass = 32 g)occupies a volume of 12 liters when its temperature is 20°C and its pressure is 1 atm.Using R = 0.082 liter.atm/mole.K, calculate the mass of the oxygen:
A)6.4 g
B)11 g
C)16 g
D)32 g
E)64 g

Accepted Answer

Using the ideal gas law PV = nRT, where P = 1 atm, V = 12 L, R = 0.082 L·atm/mol·K, and T = 273 + 20 = 293 K, we find n = PV/RT = (1*12)/(0.082*293) ≈ 0.5 moles. The mass is moles * molar mass = 0.5 * 32 = 16 g.

Question 12

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? A)1 * 10⁵ Pa and 10cm³ B)3 * 10⁵ Pa and 6 cm³ C)4 * 10⁵ Pa and 4 cm³ D)6 * 10⁵ Pa and 2 cm³ E)8 * 10⁵ Pa and 2 cm³

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 13

Air enters a hot-air furnace at 7°C and leaves at 77°C.If the pressure does not change each entering cubic meter of air expands to: A)0.80 m³ B)1.25 m³ C)1.9 m³ D)7.0 m³ E)11 m³

Accepted Answer

We can use the formula for isobaric expansion:

V2/V1 = T2/T1

where V = volume and T = temperature

Since the pressure does not change, we can assume that the initial and final volumes are equal:

V2 = V1

We can solve for V1/V2 using the temperatures:

V1/V2 = T1/T2 = (7 + 273)/(77 + 273) = 280/350 = 4/5

Therefore,

V2 = (4/5)V1

Each cubic meter of air expands to 1 + 4/5 = 1.25 m3. Therefore, the answer is choice B.

Question 14

How many molecules are in a sample of 10^-³ mol? A)6.02 x 10¹⁸ B)6.02 x 10²⁰ C)6.02 x 10²² D)6.02 x 10²³ E)cannot tell without knowing the molecular mass

Accepted Answer

The sample contains 10 moles of ^-³. The prefix "S" indicates the number of atoms of sulfur in each molecule, which is 1 in this case. The prefix "U" indicates the number of atoms of uranium in each molecule, which is also 1. The prefix "P" indicates the number of atoms of phosphorus in each molecule, which is 1 in the first molecule and 13 in the second molecule. Therefore, the total number of molecules in the sample is (10 x Avogadro's number) = 6.02 x 10^20.

Question 15

A sample of an ideal gas is compressed by a piston from 10 m³ to 5 m³ and simultaneously cooled from 273°C to 0°C.As a result there is: A)an increase in pressure B)a decrease in pressure C)a decrease in density D)no change in volume E)an increase in density

Accepted Answer

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

Question 16

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: A)21 m B)0.76 m C)4.9 m D)10 m E)0.99 m

Accepted Answer

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

Question 17

273 cm³ of an ideal gas is at 0°C.It is heated at constant pressure to 10°C.It will now occupy: A)263 cm³ B)273 cm³ C)278 cm³ D)283 cm³ E)293 cm³

Accepted Answer

The volume of an ideal gas at a constant pressure is directly proportional to its temperature in Kelvin. The initial temperature is 0°C, which is 273K, and the final temperature is 10°C, which is 283K. Since the volume is directly proportional to the temperature, increasing the temperature from 273K to 283K would also increase the volume by the same factor. Therefore, the volume increases to 283 cm³.

Question 18

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 atmospheric pressure is: A)41 B)450 C)2.4 *10²⁵ D)2.7*10²⁶ E)5.4 * 10²⁶

Accepted Answer

Using the ideal gas law, we can calculate the number of air molecules in a 1 m³ volume:
n = (P V) / (R T)
n = (1 atm * 1 m³) / (8.2 J/(mol K) * 300 K)

Convert pressure and volume to SI units:
n = (101325 Pa * 1 m³) / (8.2 J/(mol K) * 300 K)

Convert from Pa to atm and simplify:
n = (101325 Pa / 1013250 Pa/atm * 1 atm * 1 m³) / (8.2 J/(mol K) * 300 K)
n = 1.23 * 10²⁵ / (8.2 J/(mol K) * 300 K)

Substitute the given values and simplify:
n = 1.23 * 10²⁵ / (8.2 * 300 * 10⁻³ * 6.02 * 10²³)
n ≈ 2.4 * 10²⁵

Therefore, the closest choice is C, 2.4 x 10^25.

Question 19

In order that a single process be both isothermal and occur at constant pressure: A)one must use an ideal gas B)such a process is impossible C)a change of phase is essential D)one may use any real gas such as N₂ E)one must use a solid

Accepted Answer

A change of phase, such as a gas turning into a liquid or a solid, is necessary for a process to be both isothermal (occurring at constant temperature) and occur at constant pressure. This is because during a change of phase, the energy (in the form of heat) is used to break or form intermolecular attractions, rather than to increase the temperature or do work. As a result, the pressure remains constant during the phase change.

Question 20

The pressures p and volumes V of the five ideal gases, with the same number of molecules, are given below.Which has the highest temperature? A)p = 1*10⁵ Pa and V = 10cm³ B)p = 3 *10⁵ Pa and V = 6 cm³ C)p = 4 * 10⁵ Pa and V = 4 cm³ D)p = 6 * 10⁵ Pa and V = 2 cm³ E)p = 8 *10⁵ Pa and V = 2 cm³

Accepted Answer

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

Quiz 19: The Kinetic Theory of Gases

Evidence that a gas consists mostly of empty space is the fact that:

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:

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 in what property of the gas?

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

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

Platinum has a molar mass of 195 g/mol.If you have a ring that contains 2.3 g of platinum, how many moles does it contain?

Avogadro's number is:

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

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

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?

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

How many molecules are in a sample of 10^-³ mol?

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

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°C.It is heated at constant pressure to 10°C.It will now occupy:

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 atmospheric pressure is:

In order that a single process be both isothermal and occur at constant pressure:

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