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Exercises · 12.2

Q.Molar volume is the volume occupied by 1 mol1\ \text{mol} of any (ideal) gas at standard temperature and pressure (STP: 11 atmospheric pressure, 0 ∘C0\ ^\circ\text{C}). Show that it is 22.422.4 litres.

Yanam CbseNCERTSubjective· 2mImportance★★★★★est
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Using the ideal gas law PV=nRTPV = nRT at STP (P=1 atmP = 1\ \text{atm}, T=273.15 KT = 273.15\ \text{K}) with n=1 moln = 1\ \text{mol}, the molar volume comes out to 22.4 L22.4\ \text{L} — a direct consequence of Avogadro’s hypothesis that equal volumes of gases contain equal numbers of molecules.

The idea is beautifully simple. Avogadro’s hypothesis says that at the same temperature and pressure, equal volumes of all gases contain the same number of molecules. So if we can find the volume occupied by one mole of any ideal gas at a fixed reference condition (STP), that volume must be universal. The ideal gas law is the tool that lets us calculate it.

Let’s walk through it.

  1. State the ideal gas law

    The equation is PV=nRTPV = nRT, where

    PP = pressure, VV = volume, nn = number of moles, RR = universal gas constant, TT = absolute temperature.

  2. Plug in the STP conditions

    At STP:

    P=1 atmP = 1\ \text{atm} (exactly, by definition)

    T=0 ∘C=273.15 KT = 0\ ^\circ\text{C} = 273.15\ \text{K}

    n=1 moln = 1\ \text{mol} (we want the volume for one mole)

    R=0.0821 L⋅atmmol⋅KR = 0.0821\ \frac{\text{L·atm}}{\text{mol·K}} — this is the value of the gas constant in units that match litres and atmospheres.

    Tip

    Choosing the right units for RR is crucial. If you use R=8.314 J/(mol⋅K)R = 8.314\ \text{J/(mol·K)}, you’d get volume in cubic metres, which then needs conversion to litres. The value 0.0821 L⋅atm/(mol⋅K)0.0821\ \text{L·atm/(mol·K)} directly gives litres when pressure is in atm.

  3. Solve for VV

    Rearranging:

V=nRTP=(1 mol)×(0.0821 L⋅atmmol⋅K)×(273.15 K)1 atmV = \frac{nRT}{P} = \frac{(1\ \text{mol}) \times (0.0821\ \frac{\text{L·atm}}{\text{mol·K}}) \times (273.15\ \text{K})}{1\ \text{atm}}

  1. Do the multiplication First, 0.0821×273.150.0821 \times 273.15:

0.0821×273.15=22.414…0.0821 \times 273.15 = 22.414\ldots

(You can do this roughly: 0.082×273≈22.40.082 \times 273 \approx 22.4, and the exact product is 22.41422.414.)

So V=22.414 LV = 22.414\ \text{L}.

  1. Round to the familiar value To three significant figures, 22.414 L22.414\ \text{L} rounds to 22.4 L22.4\ \text{L}. That’s the standard molar volume quoted in textbooks.
Watch out

A common mistake is to use T=0 KT = 0\ \text{K} or forget to convert Celsius to Kelvin. Also, STP is sometimes defined with P=1 barP = 1\ \text{bar} instead of 1 atm1\ \text{atm} — that gives a slightly different value (22.7 L22.7\ \text{L}). For Indian exams, STP almost always means 1 atm1\ \text{atm} and 0 ∘C0\ ^\circ\text{C}, so stick with 22.4 L22.4\ \text{L}.

Note

This result is independent of the gas — whether it’s oxygen, nitrogen, or hydrogen — because the ideal gas law treats all gases identically. Real gases deviate slightly at STP, but the ideal approximation is excellent for most purposes.

✓Final answer

The molar volume at STP is 22.4 litres\boxed{22.4\ \text{litres}}.

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