Ideal gas equation: when do I use R and when k, and why must T be in kelvin?

7 answers & replies

• [+][–]Tanvi Deshmukh12thSample· 23d ago

  OKAY so!! $n$ is number of moles, $N$ is the number of molecules. They are linked by $N = nN_A$. That's why

  $nRT = Nk_BT \Rightarrow R = N_Ak_B$

  R is "per mole", $k_B$ is "per molecule". Use $PV = nRT$ whenever the question gives moles or mass/molar mass. Use $PV = Nk_BT$ when it talks about number of molecules or number density $N/V$. Values: $R = 8.314$ J mol⁻¹ K⁻¹, $k_B = 1.38\times10^{-23}$ J K⁻¹.

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• [+][–]Vivek Mishra12thSample· 23d ago

  Your mistake is only one thing: Celsius. In all gas laws temperature must be in kelvin, because the law is proportional to T from absolute zero. 27 °C is 300 K and 54 °C is 327 K. So

  $V_2 = V_1\frac{T_2}{T_1} = 2\times\frac{327}{300} = 2.18 \text{ L}$

  Volume will become double only when the kelvin temperature becomes double (300 K to 600 K, that is 327 °C). 0 °C does not mean zero energy, so ratio of Celsius values has no meaning.

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  • [+][–]Yash Tiwari11thSample· 22d agoAuthor

    oh wow 327 K. okay the Celsius mistake was the whole thing 😅 thanks Vivek bhai, ab kabhi nahi bhoolunga

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    • [+][–]Vivek Mishra12thSample· 22d ago

      No problem. Always add 273 first, before putting any number in the formula. Then it becomes a habit.

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• [+][–]Dev Agarwal12thSample· 23d ago

  About R: it is 8.314 L atm mol⁻¹ K⁻¹ if you use litres and atmospheres, and 0.0821 J mol⁻¹ K⁻¹ for SI. Just remember the pair and match with units of P and V. I trade small stock stuff, so I think of it like currency: you can't mix rupees and dollars in one formula, so pick the pair that fits your units and stay with it.

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  • [+][–]Neel Chatterjee12thSample· 22d ago

    Dev, you swapped them. $R = 8.314$ J mol⁻¹ K⁻¹ (SI, with P in Pa and V in m³), and $R = 0.0821$ L atm mol⁻¹ K⁻¹. A quick check: at STP one mole of ideal gas is 22.4 L at 1 atm and 273 K, so $R = \frac{1\times22.4}{273} = 0.0821$. That also shows why you can't mix: 8.314 with litres and atm will be way off.

    Yash, to get the old laws: constant $T$ and $n$ gives $PV$ = constant (Boyle), constant $P$ gives $V \propto T$ (Charles), constant $V$ gives $P \propto T$ (Gay-Lussac).

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• [+][–]Pooja Iyer12thSample· 22d ago

  The third form comes from replacing $n$ by mass over molar mass: $n = m/M$, so $PV = \frac{m}{M}RT$, and since density $\rho = m/V$ we get $P = \frac{\rho RT}{M}$. Useful when the question gives density, e.g. 'two gases at same P and T, ratio of densities is ratio of molar masses'. That shortcut gives marks in 10 seconds.

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