Q.Using the equation of state pV=nRT; show that at a given temperature density of a gas is proportional to gas pressure p.
Concept understanding — Ideal Gas Equation
The Intuition: What Does an "Ideal Gas" Even Mean?
Imagine a box full of tiny, perfectly bouncy balls — millions of them — zipping around in straight lines, never sticking to each other or to the walls. They take up no space themselves (their own volume is zero), and when they collide, they don't lose any energy. That's an ideal gas: a model where the only thing that matters is the motion of the particles.
Real gases (like air, oxygen, or helium) behave almost like this at low pressures and high temperatures. The ideal gas is a simplification that lets us predict how a gas will respond when we squeeze it, heat it, or add more of it.
The Three Laws That Came Before
Before the full equation, scientists discovered three separate patterns:
Boyle's Law — If you keep the temperature and amount of gas fixed, squeezing the gas into a smaller volume makes the pressure go up. Double the pressure, half the volume. Mathematically: P∝V1 (at constant n and T).
Charles's Law — If you keep the pressure and amount fixed, heating the gas makes it expand. Double the absolute temperature (in Kelvin), double the volume. So: V∝T (at constant n and P).
Avogadro's Law — If you keep pressure and temperature fixed, doubling the number of gas particles doubles the volume. So: V∝n (at constant P and T).
Each law holds a different variable constant. The genius move was to combine all three into one statement.
The Combined Statement: The Ideal Gas Equation
Putting the three proportionalities together:
V∝PnT
Remove the proportionality sign by introducing a constant R (the universal gas constant):
V=PnRT
Or, more familiarly:
PV=nRT
PV=nRT
That's it. One equation that tells you everything about the state of an ideal gas.
What Each Symbol Means
- P — Pressure of the gas (usually in pascals, Pa, or atmospheres, atm)
- V — Volume the gas occupies (in cubic metres, m³, or litres, L)
- n — Number of moles of gas (not number of molecules — one mole is 6.022×1023 particles)
- R — Universal gas constant. Its value depends on the units you use. The two you'll see most often:
- R=8.314J mol−1K−1 (when using SI units: Pa, m³)
- R=0.0821L atm mol−1K−1 (when using L and atm)
- T — Absolute temperature, always in kelvin (K). Never in Celsius. To convert: T(K)=T(°C)+273.15
Temperature must always be in kelvin. Using Celsius will give you a completely wrong answer — the equation is built on absolute zero as the starting point.
Why This Equation Is So Powerful
If you know any four of the five quantities (P, V, n, T, R), you can find the fifth. That means you can:
- Find how much gas is in a container by measuring pressure, volume, and temperature.
- Predict what happens to pressure when you heat a sealed can.
- Calculate the volume a gas will occupy at a different temperature and pressure.
A Simple Example
A 2.0 L container holds 0.50 moles of oxygen gas at 300 K. What is the pressure inside?
Use PV=nRT with R=0.0821L atm mol−1K−1:
P×2.0=0.50×0.0821×300
P=2.00.50×0.0821×300
P=6.16atm
Always check your units match the value of R you're using. If volume is in litres and pressure in atm, use R=0.0821. If volume is in m³ and pressure in Pa, use R=8.314.
The Big Picture
The ideal gas equation is not a law of nature — it's a model. It works beautifully for most gases under everyday conditions (room temperature, atmospheric pressure). It fails when gases are very cold (particles start sticking) or under very high pressure (particles get too close and their own volume matters). But for your exams and for building intuition, it's the single most important equation in gas behaviour.
The ideal gas equation PV = nRT is one of the most tested formulas in the NCERT/CBSE Class 11 Chemistry syllabus, and "ideal gas equation numericals class 11 chemistry" is among the highest-searched revision topics for this unit. It's also a near-guaranteed important-question type in JEE Main and NEET, often combined with molar mass or density calculations.
Derive density from the ideal gas equation by substituting n=m/M.
d=RTpM⟹d∝p at constant T (since M,R are constants).
Step 1 – Start from the ideal gas equation
pV=nRT
Step 2 – Express moles in terms of mass
For a gas of mass m and molar mass M:
n=Mm
Substituting:
pV=MmRT
Step 3 – Isolate density (d=m/V)
Rearranging:
p=Vm⋅MRT
Since d=m/V:
p=MdRT
⟹d=RTpM
Step 4 – Interpret at constant temperature
At a GIVEN (fixed) temperature T, for a specified gas, both M (molar mass) and R (universal gas constant) are constants. Therefore:
d=(RTM)p=(constant)×p
⟹d∝pat constant T
This confirms that density is directly proportional to pressure at a fixed temperature — doubling the pressure (compressing the gas into less space at the same T) doubles its density.
d=RTpM, so at constant T: d∝p
Substitute n=m/M into pV=nRT, then rearrange to isolate d=m/V. Whatever remains multiplying p (here, M/RT) is constant at fixed T, which is the proportionality.
- Forgetting that M (molar mass) is a constant PROPERTY of the specific gas, not a variable — it doesn't change during the process.
- Trying to prove d∝p without first substituting n=m/M — density doesn't appear in pV=nRT until this substitution is made.
- Sign/algebra slip when isolating d — keep d=m/V and p=dRT/M clearly separated before inverting to d=pM/RT.
- TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQQ.The variation of volume of an ideal gas with its number of moles (n) is obtained as a graph at 300 K and 1 atm pressure. What is the slope of the graph? (A) 24.6 L (B) 24.6 L mol−1 (C) 24.61 L−1 (D) 24.61 L−1mol
›Reveal solutionSolution
The graph plots volume V against number of moles n at fixed T and P. From the ideal gas law V=(PRT)n, the slope is PRT. At 300 K and 1 atm, R=0.0821 L⋅atm⋅mol−1K−1, so the slope is 24.6 L⋅mol−1. The correct option is (B).
Concept & Intuition
The ideal gas law, PV=nRT, relates pressure P, volume V, number of moles n, and temperature T. When P and T are held constant, V is directly proportional to n:
V=(PRT)n.
This is a linear equation of the form y=mx, where y=V, x=n, and the slope m=PRT. The slope therefore has units of volume per mole — exactly what option (B) gives. The problem asks for the slope’s numerical value and its units.
Step-by-step reasoning
- Identify the relationship At constant T=300 K and P=1 atm, the ideal gas law becomes
V=PnRT.
Since R, T, and P are constants, V is a linear function of n.
- Extract the slope The slope of the graph of V vs. n is the coefficient of n:
slope=PRT.
- Plug in the values Use the gas constant R=0.0821 L⋅atm⋅mol−1K−1 (the version with volume in liters). Then
slope=1 atm(0.0821 L⋅atm⋅mol−1K−1)(300 K)=24.63 L⋅mol−1.
Rounding to three significant figures gives 24.6 L⋅mol−1.
- Check the units The slope is volume per mole, so the correct unit is L⋅mol−1. Options (A) and (C) lack the per‑mole part; (D) has the reciprocal of the correct value and wrong units. Only (B) matches both the number and the units.
Watch outA common mistake is to forget the units. The slope is not just a number — it must carry the dimension of volume per mole. Option (A) gives the right number but wrong units (it’s a volume, not a slope). Always check what the axes represent.
TipYou can remember the slope value 24.6 L⋅mol−1 as the molar volume of an ideal gas at 300 K and 1 atm. At STP (273 K, 1 atm) it’s 22.4 L·mol⁻¹; here the higher temperature increases the volume proportionally.
✓Final answerThe correct option is (B).
ANSWER: B
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