Question 1

A sample of gas (1.3 mol)occupies ________ L at 22 °C and 3.5 atm.

Accepted Answer

Using the ideal gas law, PV=nRT:
V = (nRT)/P

Substituting given values: 
V = (1.3 mol)(0.0821 L atm mol^-1 K^-1)(295 K)/(3.5 atm)
V = 9.0 L (rounded to one decimal place)

Question 2

A sample of gas (1.9 mol)is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new pressure is 841 mm Hg and the temperature is now 26 °C. There are now ________ mol of gas in the flask.

Accepted Answer

Using the ideal gas law and the given conditions, we can calculate the moles of gas after the addition. Initially, we have $P_1 = 697 \, \text{mm Hg}$, $T_1 = 21 \, °C = 294 \, K$, and $n_1 = 1.9 \, \text{mol}$. After adding more gas, $P_2 = 841 \, \text{mm Hg}$ and $T_2 = 26 \, °C = 299 \, K$. We can use the combined gas law, rearranged to solve for $n_2$: $\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}$. Assuming the volume $V$ remains constant and canceling it out, we get $n_2 = \frac{P_2T_1n_1}{P_1T_2}$. Plugging in the values, $n_2 = \frac{841 \times 294 \times 1.9}{697 \times 299}$, which calculates to approximately 2.3 mol.

Question 3

At a temperature of ________ °C, 0.444 mol of CO gas occupies 11.8 L at 839 torr.

Accepted Answer

The answer of At a temperature of ________ °C, 0.444...

Question 4

The density of HCN is ________ g/L at STP.

Accepted Answer

The answer of The density of HCN is ________ g/L...

Question 5

A balloon originally had a volume of 4.39 L at 44 °C and a pressure of 729 torr. The balloon must be cooled to ________ °C to reduce its volume to 3.99 L (at constant pressure).

Accepted Answer

Using the combined gas law:

(P1V1/T1)=(P2V2/T2)

We know:

P1 = 729 torr 
V1 = 4.39 L 
T1 = 44 + 273.15 = 317.15 K

and

P2 = 729 torr 
V2 = 3.99 L

We need to solve for T2:

T2 = (P2V1T1)/(P1V2) = (729 torr x 4.39 L x 317.15 K) / (729 torr x 3.99 L)

T2 = 332.15 K or 59 °C

Therefore, the balloon must be cooled from 44 °C to 59 °C to reduce its volume to 3.99 L.

The closest choice is 15.1, which is Choice B.

Question 6

A 0.325 L flask filled with gas at 0.851 atm and 19 °C contains ________ mol of gas.

Accepted Answer

The answer of A 0.325 L flask filled with gas...

Question 7

If 3.21 mol of a gas occupies 56.2 L at 44 °C and 793 torr, 4.73 mol of this gas occupies ________ L under these conditions.

Accepted Answer

Using the ideal gas law, $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is temperature. Since the conditions (temperature and pressure) are the same for both amounts of gas, the volume is directly proportional to the number of moles. Therefore, we can use the ratio $\frac{3.21 \, \text{mol}}{56.2 \, \text{L}} = \frac{4.73 \, \text{mol}}{x \, \text{L}}$. Solving for $x$ gives $x = \frac{4.73 \, \text{mol} \times 56.2 \, \text{L}}{3.21 \, \text{mol}} = 82.8 \, \text{L}$.

Question 8

How many moles of gas are there in a 36.3 L container at 25.2 °C and 570.3 mm Hg?

Accepted Answer

The answer of How many moles of gas are there...

Question 9

The pressure exerted by 1.0 mol of gas in a 13 L flask at 22 °C is ________ kPa.

Accepted Answer

We can use the ideal gas law to solve for the pressure:

PV = nRT

where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature in Kelvin.

Rearranging, we get:

P = nRT/V

Plugging in the values given:

P = (1.0 mol)(8.31 J/mol*K)(295 K)/(13 L)

Converting units, we get:

P = 190 kPa

Therefore, the answer is choice B, 190.

Question 10

How many moles of gas are there in a 53.7 L container at 24.6 °C and 845.9 torr?

Accepted Answer

Using the ideal gas law, PV = nRT, where P is pressure in atm (845.9 torr = 1.112 atm), V is volume in liters (53.7 L), R is the ideal gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin (24.6 °C = 297.75 K), we can solve for n (moles of gas): n = PV/RT = (1.112 atm * 53.7 L) / (0.0821 L·atm/mol·K * 297.75 K) ≈ 2.45 moles.

Question 11

A fixed amount of gas at 25.0 °C occupies a volume of 8.66 L when the pressure is 629 torr. Use Charles's law to calculate the volume (L)the gas will occupy when the temperature is increased to 112 °C while maintaining the pressure at 629 torr.

Accepted Answer

According to Charles's law, at constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin. Therefore, we need to convert the temperatures to Kelvin: 25.0 °C = 298 K and 112 °C = 385 K. Using 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, we can solve for V2:
V2 = (V1 × T2) / T1
V2 = (8.66 L × 385 K) / 298 K
V2 = 11.2 L (rounded to two significant figures)

Question 12

A gas originally at 27 °C and 1.00 atm pressure in a 3.3 L flask is cooled at constant pressure until the temperature is 11 °C. The new volume of the gas is ________ L.

Accepted Answer

The volume of a gas is directly proportional to its temperature in Kelvin. Using the formula V1/T1 = V2/T2, where T is in Kelvin, convert the temperatures from Celsius to Kelvin (27°C = 300K, 11°C = 284K), and solve for V2. V2 = V1 * (T2/T1) = 3.3 * (284/300) = 3.1 L.

Question 13

The volume of 1.26 mol of gas at 69.4 kPa and 25.30 °C is ________ L.

Accepted Answer

Using the ideal gas law, $PV = nRT$, where $P$ is the pressure in kPa, $V$ is the volume in liters, $n$ is the number of moles, $R$ is the ideal gas constant (8.314 L·kPa/(mol·K)), and $T$ is the temperature in Kelvin. Convert 25.30°C to Kelvin: 25.30 + 273.15 = 298.45 K. Solve for $V$: $V = \frac{nRT}{P} = \frac{1.26 \times 8.314 \times 298.45}{69.4}$ ≈ 45.1 L.

Question 14

A helium balloon is filled to a volume of 27.7 L at 300 K. What will the volume of the balloon become if the balloon is heated to raise the temperature to 392 K?

Accepted Answer

Using Charles's Law, we know that the volume of a gas is directly proportional to its temperature, while pressure and amount of gas are constant. Thus, we can set up the following equation:

(V1/T1) = (V2/T2)

where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume (what we are trying to find), and T2 is the final temperature. Plugging in the values we know:

(27.7 L / 300 K) = (V2 / 392 K)

Solving for V2:

V2 = (27.7 L / 300 K) * 392 K = 36.2 L

So the balloon will expand to a volume of 36.2 L when heated to 392 K. Therefore, the answer is Choice A.

Question 15

The density of nitric oxide (NO)gas at 0.970 atm and 34.9 °C is ________ g/L.

Accepted Answer

We can use the ideal gas law to solve for density: PV=nRT. We need to solve for n/V, which is the molar density, and then multiply by the molar mass of NO (30.01 g/mol) to get the density in g/L. Rearranging the ideal gas law, we get: n/V = P/RT. Plugging in the given values, we get:

n/V = (0.970 atm)/(0.08206 L·atm/mol·K·(34.9+273) K)
n/V = 0.0395 mol/L

Now we can find the density in g/L:

density = (0.0395 mol/L)(30.01 g/mol) = 1.18 g/L

Rounding to the appropriate number of significant figures, we get 1.15 g/L, which corresponds to choice C.

Question 16

A sample of He gas (2.35 mol)occupies 57.9 L at 300.0 K and 1.00 atm. The volume of this sample is ________ L at 469 K and 1.00 atm.

Accepted Answer

The volume of the gas at the new conditions can be found using Charles's Law, which states that $V_1/T_1 = V_2/T_2$ when the pressure is constant. Given $V_1 = 57.9$ L, $T_1 = 300.0$ K, $T_2 = 469$ K, and $P = 1.00$ atm for both conditions, solving for $V_2$ gives $V_2 = V_1 \times (T_2/T_1) = 57.9 \times (469/300) = 90.5$ L.

Question 17

The volume of 0.43 mol of an ideal gas at 365 torr and 97 °C is ________ L.

Accepted Answer

The answer of The volume of 0.43 mol of an...

Question 18

The volume occupied by 2.3 mol of gas at 35 °C and 2.0 atm pressure is ________ L.

Accepted Answer

Using the ideal gas law, PV = nRT, where P = 2.0 atm, V = volume, n = 2.3 mol, R = 0.0821 L·atm/mol·K, and T = 308 K (35 °C + 273). Solving for V gives approximately 29 L.

Question 19

A sample of gas (24.2 g)initially at 4.00 atm was compressed from 8.00 L to 2.00 L at constant temperature. After the compression, the gas pressure was ________ atm.

Accepted Answer

According to Boyle's Law, P1V1 = P2V2. We can rearrange this equation to solve for the final pressure: P2 = (P1V1)/V2. Plugging in the given values, we get P2 = (4.00 atm x 8.00 L)/2.00 L = 16.0 atm. Therefore, the gas pressure after compression is 16.0 atm, which is answer choice E.

Question 20

A sample of an ideal gas (3.00 L)in a closed container at 25.0 °C and 76.0 torr is heated to 270 °C. The pressure of the gas at this temperature is ________ torr.

Accepted Answer

The answer of A sample of an ideal gas (3.00...

Quiz 10: Gases

A sample of gas (1.3 mol) occupies ________ L at 22 °C and 3.5 atm.

A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new pressure is 841 mm Hg and the temperature is now 26 °C. There are now ________ mol of gas in the flask.

At a temperature of ________ °C, 0.444 mol of CO gas occupies 11.8 L at 839 torr.

The density of HCN is ________ g/L at STP.

A balloon originally had a volume of 4.39 L at 44 °C and a pressure of 729 torr. The balloon must be cooled to ________ °C to reduce its volume to 3.99 L (at constant pressure) .

A 0.325 L flask filled with gas at 0.851 atm and 19 °C contains ________ mol of gas.

If 3.21 mol of a gas occupies 56.2 L at 44 °C and 793 torr, 4.73 mol of this gas occupies ________ L under these conditions.

How many moles of gas are there in a 36.3 L container at 25.2 °C and 570.3 mm Hg?

The pressure exerted by 1.0 mol of gas in a 13 L flask at 22 °C is ________ kPa.

How many moles of gas are there in a 53.7 L container at 24.6 °C and 845.9 torr?

A fixed amount of gas at 25.0 °C occupies a volume of 8.66 L when the pressure is 629 torr. Use Charles's law to calculate the volume (L) the gas will occupy when the temperature is increased to 112 °C while maintaining the pressure at 629 torr.

A gas originally at 27 °C and 1.00 atm pressure in a 3.3 L flask is cooled at constant pressure until the temperature is 11 °C. The new volume of the gas is ________ L.

The volume of 1.26 mol of gas at 69.4 kPa and 25.30 °C is ________ L.

A helium balloon is filled to a volume of 27.7 L at 300 K. What will the volume of the balloon become if the balloon is heated to raise the temperature to 392 K?

The density of nitric oxide (NO) gas at 0.970 atm and 34.9 °C is ________ g/L.

A sample of He gas (2.35 mol) occupies 57.9 L at 300.0 K and 1.00 atm. The volume of this sample is ________ L at 469 K and 1.00 atm.

The volume of 0.43 mol of an ideal gas at 365 torr and 97 °C is ________ L.

The volume occupied by 2.3 mol of gas at 35 °C and 2.0 atm pressure is ________ L.

A sample of gas (24.2 g) initially at 4.00 atm was compressed from 8.00 L to 2.00 L at constant temperature. After the compression, the gas pressure was ________ atm.

A sample of an ideal gas (3.00 L) in a closed container at 25.0 °C and 76.0 torr is heated to 270 °C. The pressure of the gas at this temperature is ________ torr.