NCERT Solutions for Class 11 Chemistry Chapter 5 – States of Matter
Page No 152:
Question 5.1:
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?
Answer:
Given,
Initial pressure, p1 = 1 bar
Initial volume, V1 = 500 dm3
Final volume, V2 = 200 dm3
Since the temperature remains constant, the final pressure (p2) can be calculated using Boyle’s law.
According to Boyle’s law,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5257/NS_6_11_08_Utpal_11_Chemisry_5_23_html_4cf84164.gif)
Therefore, the minimum pressure required is 2.5 bar.
Page No 153:
Question 5.2:
A vessel of 120 mL capacity contains a certain 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:
Given,
Initial pressure, p1 = 1.2 bar
Initial volume, V1 = 120 mL
Final volume, V2 = 180 mL
Since the temperature remains constant, the final pressure (p2) can be calculated using Boyle’s law.
According to Boyle’s law,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5258/NS_6_11_08_Utpal_11_Chemisry_5_23_html_80c54af.gif)
Therefore, the pressure would be 0.8 bar.
Question 5.3:
Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressurep.
Answer:
The equation of state is given by,
pV = nRT ……….. (i)
Where,
p → Pressure of gas
V → Volume of gas
n→ Number of moles of gas
R → Gas constant
T → Temperature of gas
From equation (i) we have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5259/NS_6_11_08_Utpal_11_Chemisry_5_23_html_460ea316.gif)
Replacing n with
, we have
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5259/NS_6_11_08_Utpal_11_Chemisry_5_23_html_21322174.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5259/NS_6_11_08_Utpal_11_Chemisry_5_23_html_2913a397.gif)
Where,
m → Mass of gas
M → Molar mass of gas
But,
(d = density of gas)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5259/NS_6_11_08_Utpal_11_Chemisry_5_23_html_67874527.gif)
Thus, from equation (ii), we have
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5259/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m16548c6.gif)
Molar mass (M) of a gas is always constant and therefore, at constant temperature
= constant.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5259/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m3c5258a6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5259/NS_6_11_08_Utpal_11_Chemisry_5_23_html_3873ce9f.gif)
Hence, at a given temperature, the density (d) of gas is proportional to its pressure (p)
Question 5.4:
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:
Density (d) of the substance at temperature (T) can be given by the expression,
d = ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5260/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m1336791e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5260/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m1336791e.gif)
Now, density of oxide (d1) is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5260/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m11a92ab2.gif)
Where, M1 and p1 are the mass and pressure of the oxide respectively.
Density of dinitrogen gas (d2) is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5260/NS_6_11_08_Utpal_11_Chemisry_5_23_html_651b8fe1.gif)
Where, M2 and p2 are the mass and pressure of the oxide respectively.
According to the given question,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5260/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m5a1c7184.gif)
Molecular mass of nitrogen, M2 = 28 g/mol
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5260/NS_6_11_08_Utpal_11_Chemisry_5_23_html_3a53da1f.gif)
Hence, the molecular mass of the oxide is 70 g/mol.
Question 5.5:
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:
For ideal gas A, the ideal gas equation is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5261/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m7c3aa7d3.gif)
Where, pA and nA represent the pressure and number of moles of gas A.
For ideal gas B, the ideal gas equation is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5261/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m5ce8f5cb.gif)
Where, pB and nB represent the pressure and number of moles of gas B.
[V and T are constants for gases A and B]
From equation (i), we have
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5261/NS_6_11_08_Utpal_11_Chemisry_5_23_html_40e6a592.gif)
From equation (ii), we have
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5261/NS_6_11_08_Utpal_11_Chemisry_5_23_html_74fe9efe.gif)
Where, MA and MB are the molecular masses of gases A and B respectively.
Now, from equations (iii) and (iv), we have
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5261/NS_6_11_08_Utpal_11_Chemisry_5_23_html_38127f6c.gif)
Given,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5261/NS_6_11_08_Utpal_11_Chemisry_5_23_html_133cb186.gif)
(Since total pressure is 3 bar)
Substituting these values in equation (v), we have
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5261/NS_6_11_08_Utpal_11_Chemisry_5_23_html_a7fa699.gif)
Thus, a relationship between the molecular masses of A and B is given by
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5261/NS_6_11_08_Utpal_11_Chemisry_5_23_html_1ef781e4.gif)
Question 5.6:
The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminum reacts?
Answer:
The reaction of aluminium with caustic soda can be represented as:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5262/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m60c1ea11.gif)
At STP (273.15 K and 1 atm), 54 g (2 × 27 g) of Al gives 3 × 22400 mL of H2..
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5262/NS_6_11_08_Utpal_11_Chemisry_5_23_html_4dd19828.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5262/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m37a37bc4.gif)
At STP,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5262/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m5b4dfff0.gif)
Let the volume of dihydrogen be
at p2 = 0.987 atm (since 1 bar = 0.987 atm) and T2 = 20°C = (273.15 + 20) K = 293.15 K..
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5262/NS_6_11_08_Utpal_11_Chemisry_5_23_html_3d617c0f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5262/NS_6_11_08_Utpal_11_Chemisry_5_23_html_2f6b3cdd.gif)
Therefore, 203 mL of dihydrogen will be released.
Question 5.7:
What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27 °C ?
Answer:
It is known that,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5263/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m3fcd96c4.gif)
For methane (CH4),
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5263/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m5869b1b8.gif)
For carbon dioxide (CO2),
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5263/NS_6_11_08_Utpal_11_Chemisry_5_23_html_34d1b1b7.gif)
Total pressure exerted by the mixture can be obtained as:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5263/NS_6_11_08_Utpal_11_Chemisry_5_23_html_9f00a4e.gif)
Hence, the total pressure exerted by the mixture is 8.314 × 104 Pa.
Question 5.8:
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:
Let the partial pressure of H2 in the vessel be
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5264/NS_6_11_08_Utpal_11_Chemisry_5_23_html_150f4bb9.gif)
Now,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5264/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m74b3abb9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5264/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m36c3e22d.gif)
It is known that,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5264/NS_6_11_08_Utpal_11_Chemisry_5_23_html_2671227e.gif)
Now, let the partial pressure of O2 in the vessel be
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5264/NS_6_11_08_Utpal_11_Chemisry_5_23_html_669e073d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5264/NS_6_11_08_Utpal_11_Chemisry_5_23_html_52945942.gif)
p1V1=p2V2⇒p2=p1V1V2⇒pO2=0.7×2.01= 1.4 bar
Total pressure of the gas mixture in the vessel can be obtained as:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5264/NS_6_11_08_Utpal_11_Chemisry_5_23_html_64f3e8ed.gif)
Hence, the total pressure of the gaseous mixture in the vessel is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5264/NS_6_11_08_Utpal_11_Chemisry_5_23_html_4799f56f.gif)
Question 5.9:
Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be its density at STP?
Answer:
Given,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5265/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m82b2a58.gif)
The density (d2) of the gas at STP can be calculated using the equation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5265/NS_6_11_08_Utpal_11_Chemisry_5_23_html_7b9f0a16.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5265/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m5222ed30.gif)
Hence, the density of the gas at STP will be 3 g dm–3.
Question 5.10:
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:
Given,
p = 0.1 bar
V = 34.05 mL = 34.05 × 10–3 L = 34.05 × 10–3 dm3
R = 0.083 bar dm3 K–1 mol–1
T = 546°C = (546 + 273) K = 819 K
The number of moles (n) can be calculated using the ideal gas equation as:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5266/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m2b1a9d02.gif)
Therefore, molar mass of phosphorus
= 1247.5 g mol–1
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5266/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m4b9c1f7b.gif)
Hence, the molar mass of phosphorus is 1247.5 g mol–1.
Question 5.11:
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:
Let the volume of the round bottomed flask be V.
Then, the volume of air inside the flask at 27° C is V.
Now,
V1 = V
T1 = 27°C = 300 K
V2 =?
T2 = 477° C = 750 K
According to Charles’s law,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5267/NS_6_11_08_Utpal_11_Chemisry_5_23_html_55f6c21c.gif)
Therefore, volume of air expelled out = 2.5 V – V = 1.5 V
Hence, fraction of air expelled out ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5267/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m27b50666.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5267/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m6a145dad.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5267/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m27b50666.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5267/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m6a145dad.gif)
Question 5.12:
Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar.
(R = 0.083 bar dm3 K–1 mol–1).
Answer:
Given,
n = 4.0 mol
V = 5 dm3
p = 3.32 bar
R = 0.083 bar dm3 K–1 mol–1
The temperature (T) can be calculated using the ideal gas equation as:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5268/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m6422dc7c.gif)
Hence, the required temperature is 50 K.
Question 5.13:
Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Answer:
Molar mass of dinitrogen (N2) = 28 g mol–1
Thus, 1.4 g of ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5269/NS_6_11_08_Utpal_11_Chemisry_5_23_html_76efd8e2.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5269/NS_6_11_08_Utpal_11_Chemisry_5_23_html_76efd8e2.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5269/NS_6_11_08_Utpal_11_Chemisry_5_23_html_mfac9457.gif)
Now, 1 molecule of
contains 14 electrons.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5269/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m590fe45a.gif)
Therefore, 3.01 × 1023 molecules of N2 contains = 1.4 × 3.01 × 1023
= 4.214 × 1023 electrons
Question 5.14:
How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second?
Answer:
Avogadro number = 6.02 × 1023
Thus, time required
= ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5270/NS_6_11_08_Utpal_11_Chemisry_5_23_html_62576bd8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5270/NS_6_11_08_Utpal_11_Chemisry_5_23_html_62576bd8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5270/NS_6_11_08_Utpal_11_Chemisry_5_23_html_3d09c7f9.gif)
Hence, the time taken would be
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5270/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m6f439770.gif)
Question 5.15:
Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K–1 mol–1.
Answer:
Given,
Mass of dioxygen (O2) = 8 g
Thus, number of moles of ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5271/NS_6_11_08_Utpal_11_Chemisry_5_23_html_13c391b2.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5271/NS_6_11_08_Utpal_11_Chemisry_5_23_html_13c391b2.gif)
Mass of dihydrogen (H2) = 4 g
Thus, number of moles of ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5271/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m29446043.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5271/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m29446043.gif)
Therefore, total number of moles in the mixture = 0.25 + 2 = 2.25 mole
Given,
V = 1 dm3
n = 2.25 mol
R = 0.083 bar dm3 K–1 mol–1
T = 27°C = 300 K
Total pressure (p) can be calculated as:
pV = nRT
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5271/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m17fdcd6a.gif)
Hence, the total pressure of the mixture is 56.025 bar.
Question 5.16:
Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, 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 dm3 K–1 mol–1).
Answer:
Given,
Radius of the balloon, r = 10 m
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5272/NS_6_11_08_Utpal_11_Chemisry_5_23_html_4dd19828.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5272/NS_6_11_08_Utpal_11_Chemisry_5_23_html_1b4aae08.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5272/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m251b285.gif)
Thus, the volume of the displaced air is 4190.5 m3.
Given,
Density of air = 1.2 kg m–3
Then, mass of displaced air = 4190.5 × 1.2 kg
= 5028.6 kg
Now, mass of helium (m) inside the balloon is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5272/NS_6_11_08_Utpal_11_Chemisry_5_23_html_8d02aba.gif)
Now, total mass of the balloon filled with helium = (100 + 1117.5) kg
= 1217.5 kg
Hence, pay load = (5028.6 – 1217.5) kg
= 3811.1 kg
Hence, the pay load of the balloon is 3811.1 kg.
Question 5.17:
Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure.
R = 0.083 bar L K–1 mol–1.
Answer:
It is known that,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5273/NS_6_11_08_Utpal_11_Chemisry_5_23_html_3ed5cf92.gif)
Here,
m = 8.8 g
R = 0.083 bar LK–1 mol–1
T = 31.1°C = 304.1 K
M = 44 g
p = 1 bar
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5273/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m7139daa2.gif)
Hence, the volume occupied is 5.05 L.
Question 5.18:
2.9 g of a 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:
Volume (V) occupied by dihydrogen is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5274/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m4a960020.gif)
Let M be the molar mass of the unknown gas. Volume (V) occupied by the unknown gas can be calculated as:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5274/NS_6_11_08_Utpal_11_Chemisry_5_23_html_4935a791.gif)
According to the question,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5274/NS_6_11_08_Utpal_11_Chemisry_5_23_html_349f3adb.gif)
Hence, the molar mass of the gas is 40 g mol–1.
Question 5.19:
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Answer:
Let the weight of dihydrogen be 20 g and the weight of dioxygen be 80 g.
Then, the number of moles of dihydrogen,
and the number of moles of dioxygen,
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/8183/Grade%2011_Chapter%205_html_17c20b7e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/8183/Grade%2011_Chapter%205_html_7932aee7.gif)
Given,
Total pressure of the mixture, ptotal = 1 bar
Then, partial pressure of dihydrogen,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/8183/Grade%2011_Chapter%205_html_m30673aac.gif)
Hence, the partial pressure of dihydrogen is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/8183/Grade%2011_Chapter%205_html_5ba6b806.gif)
Question 5.20:
What would be the SI unit for the quantity pV2T 2/n?
Answer:
The SI unit for pressure, p is Nm–2.
The SI unit for volume, V is m3.
The SI unit for temperature, T is K.
The SI unit for the number of moles, n is mol.
Therefore, the SI unit for quantity
is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5276/NS_6_11_08_Utpal_11_Chemisry_5_23_html_1aac750f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5276/NS_6_11_08_Utpal_11_Chemisry_5_23_html_653c5d08.gif)
Question 5.21:
In terms of Charles’ law explain why –273°C is the lowest possible temperature.
Answer:
Charles’ law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/198/5277/NS_6_11_08_Utpal_11_Chemisry_5_23_html_m730a014b.jpg)
It was found that for all gases (at any given pressure), the plots of volume vs. temperature (in °C) is a straight line. If this line is extended to zero volume, then it intersects the temperature-axis at – 273°C. In other words, the volume of any gas at –273°C is zero. This is because all gases get liquefied before reaching a temperature of – 273°C. Hence, it can be concluded that – 273°C is the lowest possible temperature.
Question 5.22:
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:
Higher is the critical temperature of a gas, easier is its liquefaction. This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, intermolecular forces of attraction are stronger in the case of CO2.
Question 5.23:
Explain the physical significance of Van der Waals parameters.
Answer:
Physical significance of ‘a’:
‘a’ is a measure of the magnitude of intermolecular attractive forces within a gas.
Physical significance of ‘b’:
‘b’ is a measure of the volume of a gas molecule.