05 States of Matter Gases and Liquids
Excersice
Question:
2.9 g of gas at 95°C occupied the same volume as 0.184 g of dihydrogen at 17°C at the same pressure. What is the molar mass of the gas?
Answer:

Convert the temperatures from Celsius to Kelvin: 95°C = 95 + 273.15 = 368.15 K 17°C = 17 + 273.15 = 290.15 K

Calculate the volume of the gas at each temperature: At 368.15 K, the volume of the gas is V1 = 2.9 g / 368.15 K = 0.0079 m^{3} At 290.15 K, the volume of the gas is V2 = 0.184 g / 290.15 K = 0.00063 m^{3}

Calculate the molar mass of the gas: Molar mass = (2.9 g / 0.0079 m^{3}) / (0.184 g / 0.00063 m^{3}) = 28.7 g/mol
Question:
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Answer:
Step 1: At 0oC, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar.
Step 2: Use the ideal gas law to calculate the molecular mass of the oxide. The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Step 3: Substitute the values for P, V, n, R, and T into the ideal gas law equation and solve for the molecular mass of the oxide.
Question:
Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses?
Answer:

Pressure of 1 g of ideal gas A at 27 °C is 2 bar.

When 2 g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar.

Therefore, the pressure of 2 g of ideal gas B at 27 °C is 3 bar.

Since both gases are ideal, the ratio of pressure is directly proportional to the ratio of the moles of the two gases.

Therefore, the ratio of the molecular masses of the two gases is equal to the ratio of the moles of the two gases.

Therefore, the relationship between the molecular masses of the two gases is directly proportional to the ratio of the pressure.
Question:
Calculate the volume occupied by 8.8 g of CO_{2} at 31.1°C and 1 bar pressure. (R = 0.083 bar L K^{(−1)} mol^{(−1)})
Answer:
Step 1: Calculate the moles of CO_{2}.
Moles = 8.8 g/44.01 g/mol = 0.200 mol
Step 2: Calculate the volume of CO_{2}.
Volume = moles x R x Temperature/Pressure Volume = 0.200 mol x 0.083 bar L K^{(−1)} mol^{(−1)} x 31.1°C/1 bar Volume = 5.05 L
Question:
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Answer:

Calculate the total weight of the mixture: Total weight = 1 bar x 100 g/cm^{3} = 100 g

Calculate the weight of dihydrogen in the mixture: Weight of dihydrogen = 20% x 100 g = 20 g

Calculate the partial pressure of dihydrogen: Partial pressure of dihydrogen = (Weight of dihydrogen / Total weight) x 1 bar = (20 g / 100 g) x 1 bar = 0.2 bar
Question:
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C ?
Answer:
Step 1: Calculate the absolute pressure of the air at 1 bar.
Absolute pressure = 1 bar + atmospheric pressure = 1 bar + 1.013 bar = 2.013 bar
Step 2: Calculate the initial volume of the air at 30°C.
Initial volume = 500 dm^{3}
Step 3: Calculate the final volume of the air at 30°C.
Final volume = 200 dm^{3}
Step 4: Calculate the compression ratio.
Compression ratio = Initial volume/Final volume = 500 dm^{3}/200 dm^{3} = 2.5
Step 5: Calculate the minimum pressure required to compress the air.
Minimum pressure = Absolute pressure x Compression ratio = 2.013 bar x 2.5 = 5.03 bar
Question:
A student forgot to add the reaction mixture to the round bottomed flask at 27°C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?
Answer:

Calculate the temperature difference between 27°C and 477°C. Temperature difference = 477°C  27°C = 450°C

Calculate the volume of air expelled using the ideal gas equation. Volume of air expelled = (nRT)/P
where, n = number of moles of air R = ideal gas constant T = temperature difference (450°C) P = atmospheric pressure
 Calculate the fraction of air expelled. Fraction of air expelled = (Volume of air expelled)/(Total volume of air)
Question:
In terms of Charles’ law, explain why  273 °C is the lowest possible temperature.
Answer:
 Charles’ law states that the volume of a gas is directly proportional to its temperature in Kelvin.
 The formula for Charles’ law is V = kT, where V is the volume, T is the temperature in Kelvin, and k is a constant.
 At absolute zero (0 K), the volume of a gas is 0.
 Since 273 °C is equal to 0 Kelvin, it is the lowest possible temperature according to Charles’ law.
Question:
Density of a gas is found to be 5.46 g/dm3 at 27°C and 2 bar pressure. What will be its density at STP? A : 3.0 g dm^{3} B : 5.0 g dm^{3} C : 6.0 g dm^{3} D : 10.82 g dm^{3}
Answer:
Step 1: Determine the STP conditions. STP stands for Standard Temperature and Pressure, which is 0°C and 1 atm.
Step 2: Calculate the density of the gas at STP using the ideal gas law, PV = nRT.
Step 3: Compare the result to the answer choices. The density at STP is 3.0 g/dm^{3}, so the correct answer is A.
Question:
Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar. (R = 0.083 bar dm^{3} K^{1} mol^{1} )
Answer:
Step 1: Calculate the number of moles of gas.
Number of moles = 4.0 mol
Step 2: Calculate the volume of the gas.
Volume of the gas = 5 dm3
Step 3: Calculate the pressure of the gas.
Pressure of the gas = 3.32 bar
Step 4: Calculate the temperature of the gas using the ideal gas law.
Temperature of the gas = (3.32 bar x 5 dm^{3}) / (0.083 bar dm^{3} K^{1} mol^{1} x 4.0 mol)
Temperature of the gas = 250 K
Question:
Payload is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload, when a balloon of radius 10 m of mass 100 kg is filled with helium at 1.66 bar at 27 °C. (Density of air = 1.2 kg m^{3} and R = 0.083 bar dm^{3} K^{1} mol^{1})
Answer:
Step 1: Calculate the mass of the balloon. Mass of balloon = (4/3)πr3 × density of air = (4/3)π × 103 × 1.2 kg/m^{3} = 478.2 kg
Step 2: Calculate the mass of displaced air. Mass of displaced air = (4/3)πr3 × density of air = (4/3)π × 103 × 1.2 kg/m^{3} = 478.2 kg
Step 3: Calculate the mass of helium at 1.66 bar and 27°C. Mass of helium = Volume of helium × density of helium = (4/3)πr3 × (1.66 bar × 0.083 bar dm^{3} K^{1} mol^{1} × 273 K) = (4/3)π × 103 × 0.0045 kg/m^{3} = 5.4 kg
Step 4: Calculate the payload. Payload = Mass of helium  Mass of displaced air = 5.4 kg  478.2 kg = 472.8 kg
Question:
A vessel of 120 mL capacity contain amount of gas at 35°C and 1.2 bar pressure The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure?
Answer:
 Calculate the amount of gas in the first vessel:
Amount of gas = (Pressure x Volume) / Temperature
Amount of gas = (1.2 bar x 120 mL) / 35°C
Amount of gas = 4.34 moles
 Calculate the pressure in the second vessel:
Pressure = (Amount of gas x Temperature) / Volume
Pressure = (4.34 moles x 35°C) / 180 mL
Pressure = 1.52 bar
Question:
Using the equation of state pV=nRT, show that, at a given temperature the density of a gas is proportional to its gas pressure p.
Answer:

The equation of state pV=nRT states that the product of pressure and volume of a gas is equal to the product of the number of moles of the gas (n), the universal gas constant (R) and the temperature (T).

Rearranging the equation of state, we get V = (nRT)/p.

Since the temperature (T) and the number of moles (n) of the gas are fixed, the volume (V) is inversely proportional to the pressure (p).

Since the density (ρ) of a gas is equal to its mass divided by its volume (ρ = m/V), it follows that the density (ρ) of a gas is proportional to its pressure (p).
Question:
The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20oC and one bar will be released when 0.15 g of aluminum reacts?
Answer:
Step 1: Calculate the amount of aluminum present in 0.15 g of Drainex.
Step 2: Calculate the amount of caustic soda required to react with 0.15 g of aluminum.
Step 3: Calculate the volume of dihydrogen released when 0.15 g of aluminum reacts with the required amount of caustic soda.
Step 4: Calculate the volume of dihydrogen at 20oC and one bar.
Question:
34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?
Answer:
 Convert 34.05 mL of phosphorus vapour to grams by multiplying by the density of phosphorus vapour at 546 °C and 0.1 bar pressure.
Density = 0.0625 g/mL
34.05 mL x 0.0625 g/mL = 2.13125 g
 Find the number of moles of phosphorus vapour.
Number of moles = Mass/Molar Mass
Number of moles = 2.13125 g/Molar Mass
 Solve for the molar mass of phosphorus.
Molar Mass = Mass/Number of moles
Molar Mass = 2.13125 g/Number of moles
Question:
Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Answer:
Step 1: Calculate the molar mass of dinitrogen gas.
Molar mass of dinitrogen gas = 28.0 g/mol
Step 2: Calculate the number of moles of dinitrogen gas present in 1.4 g.
Number of moles = 1.4 g/28.0 g/mol = 0.05 moles
Step 3: Calculate the number of electrons present in 0.05 moles of dinitrogen gas.
Number of electrons = 0.05 moles x 6.022 x 10^{23} x 2 = 7.22 x 10^{23} electrons
Question:
How much time would it take to distribute one Avogadro number of wheat grains, if 10^{10} grains are distributed each second?
Answer:
Step 1: Calculate the number of grains in one Avogadro number.
One Avogadro number is equal to 6.022 x 10^{23}. Therefore, one Avogadro number of wheat grains would be 6.022 x 10^{23} grains.
Step 2: Calculate the time required to distribute one Avogadro number of wheat grains.
The time required to distribute one Avogadro number of wheat grains would be (6.022 x 10^{23})/(10^{10}) seconds, which is equal to 5.97 x 10^{20} seconds.
Question:
Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm^{3} at 27 °C. (R = 0.083 bar dm^{3} K^{(−1)} mol^{(−1)})
Answer:
 Calculate the moles of each gas:
dioxygen: 8 g / (32 g/mol) = 0.25 mol dihydrogen: 4 g / (2 g/mol) = 2 mol
 Calculate the total number of moles:
0.25 mol + 2 mol = 2.25 mol
 Calculate the total volume:
1 dm3 = 1 L = 1000 cm^{3}
 Calculate the total pressure:
P = nRT/V P = (2.25 mol) (0.083 bar dm^{3} K^{(−1)} mol^{(−1)}) (300 K) / (1000 cm^{3}) P = 0.619 bar
Question:
Critical temperature for carbon dioxide and methane are 31.1°C and −81.9°C respectively. Which of these has stronger intermolecular forces and why?
Answer:
Answer: Methane has stronger intermolecular forces because its critical temperature is lower than that of carbon dioxide. This indicates that the intermolecular forces in methane are stronger than those in carbon dioxide, since the molecules of methane must be brought closer together to reach the critical temperature.
Question:
What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of CO2 contained in a 9 dm^{3} flask at 27 °C.
Answer:
Step 1: Calculate the moles of methane (CH_{4}) and carbon dioxide (CO_{2}) present in the flask.
Moles of CH_{4} = 3.2 g / 16 g/mol = 0.2 mol Moles of CO_{2} = 4.4 g / 44 g/mol = 0.1 mol
Step 2: Calculate the total number of moles of gas present in the flask.
Total moles = 0.2 mol + 0.1 mol = 0.3 mol
Step 3: Calculate the pressure exerted by the gas mixture using the ideal gas equation.
PV = nRT P = (0.3 mol x 8.314 J/mol K x 300 K) / 9 dm^{3} P = 8.7 kPa
Question:
What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27 °C ?
Answer:
 Calculate the total moles of gas in the vessel:
Moles of H2 = 0.5 L x (0.8 bar x 1 atm/101.325 kPa) x (1 mol/22.4 L) = 0.0358 moles
Moles of O2 = 2.0 L x (0.7 bar x 1 atm/101.325 kPa) x (1 mol/22.4 L) = 0.1464 moles
Total moles = 0.0358 + 0.1464 = 0.1822 moles
 Calculate the total pressure of the gas mixture:
Total pressure = (0.0358 moles x 0.8 bar) + (0.1464 moles x 0.7 bar) = 0.1822 bar
 Calculate the pressure of the gaseous mixture in the vessel:
Pressure of the gaseous mixture = 0.1822 bar x (1L/1L) = 0.1822 bar
Question:
What would be the SI unit for the quantity pV^{2T}^{2}/n?
Answer:
Answer: joules (J)
Question:
Explain the physical significance of Van der Waals parameters.
Answer:
Step 1: Van der Waals parameters are parameters used to describe the behavior of gases and liquids.
Step 2: The Van der Waals parameters are used to describe the interactions between molecules of a gas or liquid, and how these interactions affect the physical properties of the substance.
Step 3: These parameters are related to the forces of attraction and repulsion between molecules, and are used to calculate the pressure, volume, temperature, and other properties of the substance.
Step 4: The physical significance of Van der Waals parameters is that they provide an understanding of how gases and liquids interact with each other, and how these interactions affect the physical properties of the substance.
01 Some Basic Concepts of Chemistry
02 Structure of Atom
03 Classification of Elements and Periodicity in Properties
04 Chemical Bonding and Molecular Structure
05 States of Matter Gases and Liquids
06 Thermodynamics
07 Equilibrium
08 Redox Reactions
09 Hydrogen
10 The s block elements
11 Some p block elements
12 Organic Chemistry some basic principles and techniques
13 Hydrocarbons
14 Environmental