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IUPAC STP Molar Volume at 1 bar and 273.15 K

At IUPAC STP (pressure 1 bar and temperature 273.15 K), what is the molar volume of an ideal gas, and how does it compare with the traditional STP value at 1 atm?

Subject: General Chemistry Chapter: Gases Topic: Molar Volume Stp and Rtp Answer included
iupac stp 1 bar 273.15 k molar volume 22.711 l IUPAC STP standard molar volume molar volume at STP ideal gas law gas constant R 1 bar vs 1 atm traditional STP
Accepted answer Answer included

What the keyword means

The phrase iupac stp 1 bar 273.15 k molar volume 22.711 l refers to the standard molar volume of an ideal gas at IUPAC STP, where the standard conditions are: \(T = 273.15\,\mathrm{K}\) and \(P = 1\,\mathrm{bar}\).

Goal: Find the molar volume \(V_m\) (volume per 1 mole) at IUPAC STP and explain why it differs from the older “STP at 1 atm” value.

Step 1: Start from the ideal gas equation

For an ideal gas,

\[ P \cdot V = n \cdot R \cdot T. \]

The molar volume is \(V_m = \frac{V}{n}\), so dividing both sides by \(n\) gives

\[ P \cdot V_m = R \cdot T \quad \Longrightarrow \quad V_m = \frac{R \cdot T}{P}. \]

Step 2: Use IUPAC STP values and consistent units

IUPAC STP uses \(T = 273.15\,\mathrm{K}\) and \(P = 1\,\mathrm{bar}\). Choosing the gas constant in matching units,

\[ R = 0.083145\,\mathrm{L\cdot bar\cdot mol^{-1}\cdot K^{-1}}. \]

Step 3: Compute the molar volume at IUPAC STP

\[ V_m = \frac{(0.083145\,\mathrm{L\cdot bar\cdot mol^{-1}\cdot K^{-1}})\cdot (273.15\,\mathrm{K})}{1\,\mathrm{bar}} = 22.711\,\mathrm{L\cdot mol^{-1}} \; (\text{rounded}). \]

Small differences in the last digits can appear if \(R\) is rounded differently; the standard reported value is \(22.711\,\mathrm{L\cdot mol^{-1}}\) at IUPAC STP.

Step 4: Compare with “traditional STP” at 1 atm

Many textbooks historically used STP as \(T=273.15\,\mathrm{K}\) and \(P=1\,\mathrm{atm}\), where \(1\,\mathrm{atm}=1.01325\,\mathrm{bar}\). Using the same formula,

\[ V_m(1\,\mathrm{atm}) = \frac{R \cdot T}{P} \approx \frac{(0.083145)\cdot (273.15)}{1.01325} \approx 22.414\,\mathrm{L\cdot mol^{-1}}. \]

The IUPAC value is larger because \(1\,\mathrm{bar}\) is slightly lower pressure than \(1\,\mathrm{atm}\), and at fixed \(T\), lower pressure implies larger volume for the same amount of gas.

Condition label \(T\) \(P\) Ideal-gas molar volume \(V_m\)
IUPAC STP \(273.15\,\mathrm{K}\) \(1\,\mathrm{bar}\) \(22.711\,\mathrm{L\cdot mol^{-1}}\)
Traditional STP \(273.15\,\mathrm{K}\) \(1\,\mathrm{atm}=1.01325\,\mathrm{bar}\) \(22.414\,\mathrm{L\cdot mol^{-1}}\)

Visualization: molar volume at 1 bar vs 1 atm (same temperature)

0 5 10 15 20 25 Molar volume \(V_m\) (L·mol⁻¹) at \(T=273.15\,\mathrm{K}\) IUPAC STP: \(P=1\,\mathrm{bar}\) Traditional STP: \(P=1\,\mathrm{atm}\) \(22.711\,\mathrm{L\cdot mol^{-1}}\) \(22.414\,\mathrm{L\cdot mol^{-1}}\) Lower pressure (1 bar) gives a slightly larger molar volume than 1 atm at the same temperature.
Both conditions use \(273.15\,\mathrm{K}\). The small difference in bar heights visualizes the pressure change from \(1\,\mathrm{atm}\) to \(1\,\mathrm{bar}\), which shifts the ideal-gas molar volume from about \(22.414\) to \(22.711\,\mathrm{L\cdot mol^{-1}}\).

Final result

At IUPAC STP (\(1\,\mathrm{bar}\), \(273.15\,\mathrm{K}\)), the ideal-gas molar volume is \(V_m \approx 22.711\,\mathrm{L\cdot mol^{-1}}\), and it is larger than the traditional \(1\,\mathrm{atm}\) STP value because \(1\,\mathrm{bar} < 1\,\mathrm{atm}\).

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