Question 1

A glass column is filled with mercury and inverted in a pool of mercury.The mercury column stabilizes at a height of 736 mm above the pool of mercury.What is the pressure of the atmosphere?

Accepted Answer

The height of the mercury column (736 mm) can be converted to pressure using the equation P = ρgh, where ρ is the density of mercury (13,595 kg/m³), g is the acceleration due to gravity (9.81 m/s²), and h is the height of the column in meters. 
P = (13,595 kg/m³)(9.81 m/s²)(0.736 m) = 101,325 Pa = 1 atm
Therefore, the pressure of the atmosphere can be calculated by subtracting the pressure of the mercury column from atmospheric pressure (1 atm).
Patm = 1 atm - 0.968 atm = 0.032 atm
Therefore, the best choice is D (0.968 atm).

Question 2

A flexible container is filled with 49.00 L of gas at 333 Κ.Under conditions of constant pressure and moles of gas,what is the temperature of the gas when the volume is doubled?

Accepted Answer

The relationship between volume and temperature at constant pressure and moles of gas is described by Charles's Law, which states that V1/T1 = V2/T2. Given V1 = 49.00 L and T1 = 333 K, when V2 is doubled to 98.00 L, solving for T2 gives T2 = (98.00 L * 333 K) / 49.00 L = 666 K.

Question 3

A flexible vessel is filled to a certain pressure with 93.00 L of gas.Under conditions of constant temperature and moles of gas,how does the volume of the gas change when the pressure of the gas is tripled?

Accepted Answer

According to Boyle's Law, at constant temperature and moles of gas, the pressure of a gas is inversely proportional to its volume. That is, as pressure increases, volume decreases, and vice versa. Mathematically, P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, respectively, and P2 and V2 are the final pressure and volume, respectively. If the pressure is tripled, then P2 = 3P1. Using the formula and solving for V2, we get V2 = V1(P1/P2) = V1(1/3) = 31.00 L. Therefore, the volume decreases by a factor of three.

Question 4

A gas sample is held at constant pressure.The gas occupies 3.62 L of volume when the temperature is 21.6°C.Determine the temperature at which the volume of the gas is 3.47 L.

Accepted Answer

We can use the formula V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.

Plugging in the given values, we get:

V1 = 3.62 L
T1 = 21.6°C + 273.15 = 294.75 K
V2 = 3.47 L

Solving for T2, we get:

T2 = (V2/T2) x T1
T2 = (3.62/3.47) x 294.75
T2 = 306.61 K

Rounding to the nearest whole number, we get:

T2 = 307 K

Therefore, the correct answer is (B) 283 K.

Question 5

A sample of methane,CH₄,occupies a volume of 230.0 mL at 25°C and exerts a pressure of 900.0 mmHg.If the volume of the gas is allowed to expand to 680.0 mL at 298 K,what will be the pressure of the gas?

Accepted Answer

To solve this problem, we have to use the combined gas law equation: 
P₁V₁/T₁ = P₂V₂/T₂ 
where P₁, V₁, T₁ are the initial pressure, volume, and temperature, and P₂, V₂, T₂ are the final pressure, volume, and temperature. 
We have to convert 25°C to Kelvin: 
T₁ = 25°C + 273.15 = 298.15 K 
T₂ = 298 K 
We also have to convert mmHg to atm: 
P₁ = 900.0 mmHg × (1 atm/760 mmHg) = 1.1842 atm 
V₁ = 230.0 mL = 0.2300 L 
V₂ = 680.0 mL = 0.6800 L 
Now we can substitute these values in the equation and solve for P₂: 
P₂ = (P₁V₁/T₁) × T₂/V₂ 
P₂ = (1.1842 × 0.2300/298.15) × 298/0.6800 
P₂ = 0.9138 atm 
Finally, we have to convert atm to mmHg: 
P₂ = 0.9138 atm × (760 mmHg/1 atm) = 693.9 mmHg ≈ 304.4 mmHg 
Therefore, the answer is C.

Question 6

A flexible vessel contains 53 L of gas where the pressure is 1.1 atm.What will the volume be when the gas is compressed to a pressure of 0.91 atm,the temperature remaining constant?

Accepted Answer

According to Boyle's Law, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure. The formula is P1V1 = P2V2. Substituting the given values: 1.1 atm * 53 L = 0.91 atm * V2. Solving for V2 gives V2 = (1.1 * 53) / 0.91 = 64 L.

Question 7

Which of the following is the largest value of pressure?

Accepted Answer

The answer of Which of the following is the largest...

Question 8

A gas sample is heated from −20.0°C to 57.0°C and the volume is increased from 2.00 L to 4.50 L.If the initial pressure is 0.150 atm,what is the final pressure?

Accepted Answer

Using the combined gas law (P1V1/T1 = P2V2/T2), convert temperatures to Kelvin and solve for P2. The final pressure is 0.0869 atm.

Question 9

Which of the following statements is true of Charles's law,  ?

Accepted Answer

Charles's law states that the volume of a gas sample varies directly with the absolute temperature, when pressure and amount of gas are held constant. This means that as the temperature increases, the volume of the gas sample also increases. Thus, choice B is the correct statement of Charles's law. Choices A, C, and D describe other gas laws (Boyle's law, Avogadro's law, and Gay-Lussac's law, respectively). Choice E is not a law, but rather a statement about the behavior of gases at STP (standard temperature and pressure).

Question 10

A flexible vessel contains 49.00 L of gas at a pressure of 2.81 atm.Under the conditions of constant temperature and constant number of moles of gas,what is the pressure of the gas when the volume of the vessel is quadrupled?

Accepted Answer

The pressure of the gas decreases inversely with the increase in volume when temperature and moles of gas are constant (Boyle's Law: P1V1 = P2V2). Quadrupling the volume (4 * 49.00 L) results in the pressure being one-fourth of the original (2.81 atm / 4 = 0.7025 atm, rounded to 0.702 atm).

Question 11

The local weather forecaster reports that the current barometric pressure is 29.2 inches of mercury.What is the current pressure in atmospheres?

Accepted Answer

The answer of The local weather forecaster reports that the...

Question 12

A sample of helium gas occupies 17.9 L at 23°C and 0.956 atm.What volume will it occupy at 40°C and 1.20 atm?

Accepted Answer

The answer of A sample of helium gas occupies 17.9...

Question 13

A physics experiment is conducted at a pressure of 13.8 kPa.What is this pressure in mmHg? (1 atm = 760 mm Hg = 101.3 kPa)

Accepted Answer

The answer of A physics experiment is conducted at a...

Question 14

A particular gas exerts a pressure of 5.35 × 10⁴ Pa.What is this pressure in units of atmospheres? (1 atm = 760 mm Hg = 101.3 kPa = 1.013 bar)

Accepted Answer

To convert the pressure from pascals to atmospheres, use the conversion factor 1 atm = 101.3 kPa. Given pressure is 5.35 × 10^14 Pa = 5.35 × 10^5 kPa. So, $5.35 \times 10^5 \, \text{kPa} \times \frac{1 \, \text{atm}}{101.3 \, \text{kPa}} \approx 0.528 \, \text{atm}$.

Question 15

It is possible to make a barometer using a liquid other than mercury.What would be the height (in meters)of a column of dichloromethane at a pressure of 0.790 atm,given that 0.790 atm is equal to a 0.600 m column of mercury and the densities of mercury and dichloromethane are 13.5 g/cm³ and 1.33 g/cm³,respectively.

Accepted Answer

The height of a liquid column in a barometer is inversely proportional to the density of the liquid. Given that 0.790 atm equals a 0.600 m column of mercury, to find the height of a column of dichloromethane, we use the ratio of their densities. The height of the dichloromethane column is (13.5 g/cm³ / 1.33 g/cm³) * 0.600 m = 6.09 m.

Question 16

A certain person has a body temperature of 309 K.On a cold day,what volume of air at 276 K must this person with a lung capacity of 3.1 L breathe in to fill his lungs?

Accepted Answer

The answer of A certain person has a body temperature...

Question 17

A balloon has a volume of 2.52 liters at 24.0°C.The balloon is heated to 48.0°C.Calculate the new volume of the balloon.

Accepted Answer

We can use the formula for the relationship between temperature and volume at constant pressure:

V₂ = V₁ × (T₂/T₁)

where V₁ and T₁ are the initial volume and temperature, and V₂ and T₂ are the final volume and temperature.

Plugging in the values:

V₂ = 2.52 L × (48.0°C/24.0°C) = 2.52 L × 2 = 5.04 L

Therefore, the new volume of the balloon is 5.04 L. The correct answer is B).

Question 18

A particular gas exerts a pressure of 2.97 bar.What is this pressure in units of atmospheres? (1 atm = 760 mm Hg = 101.3 kPa = 1.013 bar)

Accepted Answer

The answer of A particular gas exerts a pressure of...

Question 19

A particular gas exerts a pressure of 681 mm Hg.What is this pressure in units of bar? (1 atm = 760 mm Hg = 101.3 kPa = 1.013 bar)

Accepted Answer