Ideal Gases
Under all temperature and pressure conditions, a gas that obeys the general gas equation and other gas laws is called an ideal gas or perfect gas.
Ideal Gas equation Units
| Terms |
Symbol |
Definition |
Units |
|---|---|---|---|
| Pressure |
P |
Force per units area |
Pa or N/m2 |
| Volume |
V |
Three-dimensional space enclosed by the closed surface |
m3 |
| Amount of substances/number of moles |
n |
The ratio of the mass of the gas(m) to its molar mass(M). |
Mole |
| Ideal gas constant |
R |
Physical constant relating the average kinetic energy of an ideal gas with temperature. |
8.3144598(48) J.K-1.mol-1 |
| Temperature |
T |
Measure of heat |
K or 0C |
Explanation
The specialists have discovered that regardless of what gas study, if taken a one-mole test of that gas and put it in a similar holder and keep a steady temperature, the crucial factor is nearly the equivalent, and at lower densities, even those minuscule contrasts in the estimations additionally vanish.
Accordingly, at extremely low densities, all the natural gases will, in general, comply with one all-inclusive law called ideal gas law.
This law is described by an equation known as the Ideal gas equation: PV = nRT
It is a complicated part of gas constants; R. R-value will change when dealing with various pressure units and volumes (temperature factors are ignored because the temperature will always be in Kelvin, not Celsius when using the ideal gas equation). Only through the right R-value you will get the correct answer to the problem.
Ideal Gas Law in Class 11
In class 11, the ideal gas law is in the chapter- State of matter, where ideal gas has been discussed with its laws, properties, and questions about the same. The chapter- State of Matter has a weightage of 20 Marks.
Illustrated Examples
1. Compute the absolute pressing factor in a combination of 10 g of dioxygen and 5 g of dihydrogen kept in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K–1 mol–1.
Solution: m (Oxygen) = 10 g,
M (Oxygen) = 32 g/mol
m ( Hydrogen) = 5 g,
M (Hydrogen) = 2 g/mol
n (Amount of oxygen required) = 10/ 32 = 0.31 mol
n (Amount of hydrogen required) = 5/2 = 2.5 mol
As per the ideal gas equation,
PV = nRT,
P X 1 = (0.25 + 2) X 0.31 X 300 = 50.02 bar
Therefore, the pressure of the mixture is 50.02 bars.
2. Figure the volume involved by 8.8 g of CO2 at 31.1°C and 1 bar pressure. R = 0.083 bar
Solution: As per the ideal gas equation, PV = nRT
But n = (m)/ (M)
PV = (m/M) RT
For CO2 , M = 44 g/mol
Placing the values
1 x V = (8.8 / 44) x 0.083 x 304.1
= 5.05 L
Therefore, the volume occupied is 5.05 L.
3. Utilizing the condition of state PV = NRT regards the given temperature as gas pressure.
Solution: As per the Ideal gas equation, pV = nRT ……….. (1)
Taking (1),
p = n RT/V
n = Mass of Gas (m) /Molar Mass of gas (M)
Placing the value of n in the equation, we get
p = m RT/ MV ------------(2)
Density (ρ) = m /V ----------------(3)
Putting (3) in (2) we get
p = ρ RT / M
Or ρ = PM / RT
FAQs on Ideal Gas Law
Q: Using the equation of state PV= nRT, show that at a given temperature, density of a gas is proportional to the gas pressure P.
- A: According to ideal gas equation
- PV = nRT Or
- P= nRT/V Replacing n by m/M, we get
- P = mRT/MV Replacing m/V by d (i.e. density), we get
- P = dRT/M
- P ∝ d Hence, at a given temperature, the density of a gas is directly proportional to its pressure.
Q: The drain cleaner, Drainex contains small bits of aluminium 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 aluminium reacts?
A: The chemical equation for the reaction is
- 2 Al + 2 NaOH + H2O→2 NaAlO2 + 3H2
- i.e. 2 moles of Al give 3 moles of H2 gas.
- Moles of aluminium = 270.15g = 5.56×10−3 moles
- Moles of H2 produced= 23×5.56×10−3
- = 8.33×10−3 moles
- Volume of H2 = nRT / P
- = 8.33×10−3 x 0.0831×293= 0.203 litres
- Therefore, volume of dihydrogen released is 0.203 litres.
Q: What will be the pressure exerted by a mixture of 3.2g of methane and 4.4g of carbon dioxide contained in a 9 dm^3 flask at 27 °C?
A: Applying PV = nRT
- We get P = nRT / V
- Given: nCH4 = 3.2 / 16 mol = 0.2 mol
- nCO2 = 4.4 /44 mol = 0.1 mol
- So, PCH4 = (0.2 mol) (0.0821 dm3atm K-1 mol-1) (300 K) / (9 dm3)
- = 0.55 atm
- PCO2= (0.1 mol) (0.0821 dm3atm K-1 mol-1) (300 K) / (9 dm3)
- = 0.27 atm
- Ptotal = PCH4 + PCO2 = 0.55 + 0.27
- = 0.82 atm
- = 8.314 x 104 Pa
- Hence, the pressure exerted by a mixture is 8.314 x 104 Pa
Q : 34.05 mL of phosphorus vapour weighs 0.0625 g at 546°C and 1.0 bar pressure. What is the molar mass of phosphorus?
A: We know,
- PV=nRT
- n= PV/RT
- n= (0.1 x 34.05 x 10-3) / (0.083 x 819)
- = 5.01 x 10-5 mol
- Hence, the atomic mass of phosphorus= 0.0625/ (5.01 x10-5)
- = 1247.5 g/mol
Q: Calculate the temperature of 4.0 moles of a gas occupying 5 dm^3 at 3.32 bar (R = 0.083 bar dm^3 K^-1 mol^-1).
A: Given,
- P= 3.32 bar
- V= 5 dm3
- n= 4 mol
- R= 0.083 bar dm3 K-1 mol-1
- PV = nRT
- Or T = PV / nR = 3.32 x 5 dm3 / (4.0 mol x 0.083 bar dm3 K-1 mol-1)
- = 50 K
Q: Calculate the total pressure in a mixture of 8g of oxygen and 4g of hydrogen confined in a vessel of l dm^3 at 27°C. R = 0.083 bar dm^3 K^-1 mol^-1.
A: Molar mass of O2 = 32 g/mol
- It means, 8 g of O2 has 8/32 mol = 0.25 mol
- Molar mass of H2 = 2 g/mol
- It means, 4 g of H2 has 4/2 mol = 2 mol
- Therefore, total number of moles, n = 2 + 0.25 = 2.25 mol
- Given, V = 1dm3, T = 27°C = 300 K, R = 0.083 bar dm3 K-1 mol-1
- Applying PV = nRT, P = nRT / V = (2.25) (0.083 bar dm3 K-1 mol-1) (300 K) / (1dm3)
- = 56.025 bar
Q: Pay load is defined as the difference between the mass of the 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 dm^3 K^-1 mol^-1).
A: Radius of the balloon = 10 m
- Therefore, volume of the balloon = (4/3) π r3 = (4/3) x (22/7) x (10 m)3
- = 4190.5 m3
- Volume of He filled at 1.66 bar and 27 °C = 4290.5 m3
- To calculate the mass of He,
- PV = nRT = (w/M) RT, where M is molar mass of He i.e. 4 g per mole or 4 x 10-3 kg mol-1
- => w = MPV/RT
- = [(4 x 10-3 kg mol-1) (1.66 bar) (4190.5 m3)] / [(0.083 bar dm3 K-1 mol-1) (300K)]
- = 1117.5 kg
- Total mass of the balloon along with He = 100 + 1117.5 = 1217.5 kg
- Maximum mass of the air that can be displaced by balloon to go up = volume x density
- = 4190.5 m3 x 1.2 kg m-3 = 5028.6 kg
- Hence, pay load = 5028.6 – 1217.5 kg = 3811.1 kg
Q: Calculate the volume occupied by 8.8 g of CO2 at 31.1 °C and 1 bar pressure. R = 0.083 bar LK^-1 mol^-1
- A: No. of moles of CO2 = Given mass of CO2 / Molar mass
- = 8.8g / 44g mol-1 = 0.2 mol
- Pressure of CO2 = 1 bar
- RR = 0.083 bar dm3 K–1 mol–1
- T = 273 + 31.1 K = 304.1 K
- According to ideal gas equation,
- PV = nRT
- Therefore, V = nRT/P
- = (0.2 x 0.083 x 304.1) / 1 bar = 5.048 L
Q: The shape of the P-T curve of an ideal gas is?
A: The shape of the P-T curve of an ideal gas is a straight line
Q: The compressibility factor of an ideal gas is?
A: The compressibility factor of an ideal gas is always one
Q: The shape of a V-T curve of an ideal gas is?
A: The shape of the V-T curve for an ideal gas is a straight line
Q: In statistical mechanics, the ideal gas equation is expressed as?
A: In statistical mechanics, the ideal gas equation is given by:-
Q: What is the most common form of the ideal gas equation?
A: Most frequently used form of the ideal gas equation is PV = nRT = NkBT
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