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How much methane gas ($\text{cm}^3$) is required heat $150~\text{m}^3$ of air by $10~^\circ\text{C}$?

Please help! I thought this wouldn't be too difficult but I've been stuck for at least an hour now- so frustrating! Thank you so much in advance!

Edit: Thank you for the comment, here's what I did:

Let the specific heat capacity of air be $1.15~\frac{\text{J}}{\text{g K}}$ (this is apparently the heat capacity for some average temperature and pressure- not sure if it's reliable or not)

\begin{aligned} E&=mc\Delta T\\ &=150,000,000~\text{cm}^3 \cdot 1.15 \cdot 10\\ &=1,725,000,000~\text{J} = 1,725,000~\text{kJ} \end{aligned}

Any help as to where I go from here? It appears that I should find out how many $\text{mol}$ of $\ce{CH4}$ is needed to produce $1,725,000~\text{kJ}$ (which is the enthalpy change or $\Delta H$) and then find the volume of $\ce{CH4}$ but I have no idea how to do that- either because I am not using the right information or I'm just not very good with calculations (probably a mixture of both). Do I find the moles of $\ce{CH4}$ in the room first?

Second edit: Sorry! It's because half of them are just invalid as I got confused over which numbers to use and where. I tried using Charles' Law $V_1/T_1 = V_2/T_2$ but that didn't work, then I thought I would find the enthalpy change but my units got messed up and there were all sorts of odd numbers coming up. Plus I spent quite a bit of time finding the heat capacities and densities etc.

Third edit: I solved it! First found out the $E$ from $E=mc\Delta T$ of room, then mol of $\ce{CH4}$ needed to fill room (using knowledge that enthalpy change of $\ce{CH4}$ is $891~\text{kJ}\cdot\text{mol}^{-1}$. Lastly found out volume by $n=v/24000$.

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  • $\begingroup$ The only values you need to look up to do the calculation are the molar combustion enthalpy of methane and the volumetric heat capacity of air (actually you could even approximate the heat capacity to that of an ideal diatomic gas). Does that help? $\endgroup$ – Nicolau Saker Neto Dec 1 '13 at 12:31
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    $\begingroup$ If you've been working on it for 'at least an hour' then can you show us what you have done, equations you have used, and why you disregarded those approaches? $\endgroup$ – bobthechemist Dec 1 '13 at 13:17
  • $\begingroup$ You've solved half of the problem; you know how much heat it takes to produce the required warming. Now, all you need to know is how much methane produces that heat. Using the molar combustion enthalpy of methane, you can find out how many moles it takes. With the number of moles, I imagine all you need to do is apply $PV=nRT$ in some reasonable conditions, such as $P=1\ bar$ and $T=298\ K$. Got it? $\endgroup$ – Nicolau Saker Neto Dec 1 '13 at 20:19
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    $\begingroup$ Wait, now I see you have a problem with units. If the specific heat of air you found is given in units of $Jg^{-1}K^{-1}$, then you need to find the mass of air in $150\ m^3$. You need to know the initial temperature of the room to calculate it, though. $\endgroup$ – Nicolau Saker Neto Dec 1 '13 at 20:31
  • $\begingroup$ New edit: solved question $\endgroup$ – sonder Dec 1 '13 at 23:10
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1) \begin{aligned} E&=mc\Delta T\\ &=1.012\color{\navy}{\:\mathrm{\frac{J}{g\cdot{}^\circ{}C}}} \cdot (150\color{\navy}{\:\mathrm{m^3}} \cdot 1.2\:\mathrm{g/L}) \cdot 10\:\mathrm{^\circ{}C}\\ &=1821.6\:\mathrm{\color{\navy}{k}J} \end{aligned}

2) Find Mol $\ce{CH4}$ required to fill room

\begin{aligned} \ce{CH4} &\text{ releases }891\:\mathrm{kJ}\text{ per }\:\mathrm{mol}\\ &=1.8216\cdot\color{\navy}{10^3}\:\mathrm{kJ}/ 891\color{\navy}{\:\mathrm{\frac{kJ}{mol}}}\\ &=2.04\cdot\color{\navy}{10^3}\:\mathrm{mol} \end{aligned}

3) \begin{aligned} n&= v/24000\cdot\color{\navy}{\:\mathrm{cm^3}}\\ v&= n \cdot 24000\cdot\color{\navy}{\:\mathrm{cm^3}}\\ &=2.04\cdot\color{\navy}{10^3\:\mathrm{mol}} \cdot 24000\color{\navy}{\:\mathrm{cm^3}}\\ &=49066\cdot\color{\navy}{10^3}\:\mathrm{cm^3}\\\hline \end{aligned} (coloured items are added by Martin upon Nicolau's comment.)

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  • $\begingroup$ Note that $150\ m^3=150,000\ L$, so your answer is off by a factor of a thousand. $\endgroup$ – Nicolau Saker Neto Dec 3 '13 at 0:24
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    $\begingroup$ due to colors answer does not look boring :) $\endgroup$ – Freddy Jun 11 '14 at 10:19

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