# Applications of Thermodynamics

I am trying to solve this but I am stuck; I have watched loads of YouTube videos but still don't understand how to complete it:

A mass of $m=0.12 kg$ of air has an initial temperature of $T_1=500°C$ and pressure of $p_1=0.8 MPa$. If the air is expanded according to the law $pV^{1.2} = c$ to a final volume of $90\ litres$, determine

i) its initial volume, $V_1$

ii) its final pressure, $p_2$

iii) its final temperature. $T_2$

For air, take $R_{specific} = 287 Jkg^{-1} K^{-1}.$

I have got these equations which i believe i need to use.

$$pV = nRT$$

$$n = \frac{p V}{R T}$$

Are these the correct equations to use?

I also think the fixed amount of gas is the constant?

I have been looking at Boyle's law. And Charles' law.

Any help appreciated.

Thanks.

• You are given pv1.2 = c , so what have you done so far with that - gamma is Cv / Cp ... or is it the other way over ? Feb 7 '18 at 14:31
• pv1.2 = c is what i don't understand to well. Do i do: 0.8 x 90 x 1.2 = c? Feb 7 '18 at 14:50
• No, it means that the process is adiabatic, so you have p*V^gamma = constant with gamma = 1.2. gamma is defined as the ratio Cp/Cv. Feb 7 '18 at 14:53
• There are far better ways to learn a few equations than watching videos. Feb 7 '18 at 15:58
• What other ways? I have always just watched YouTube videos. Feb 7 '18 at 15:59

First step, is to find the value of '$n$'. The molar mass of air is $29 g/mol$, meaning that in $0.12kg$, you have $4.144mol$.

Remember to convert all your units into standard SI values:

$p_1=800,000Pa$

$n=4.144mol$

$R=8.31441JK^{-1} mol^{-1}$ - N.B. You have been suplied with the R in terms of kg, rather than mol. This can make the maths easier, but is often a source of confusion. I recommend always using the Universal gas constant, and calculating the number of mol that you have, rather than using Specific Constants.

$T_1=773.15K$

Rearranging $pV=nRT$ to give $V=\frac{nRT}{p}$ allows you to calculate the initial volume.

i) $V_1=\frac{4.144*8.31441*773.15}{800,000}=0.03339m^3$ You can verify this via Wolfram|Alpha

Next, the fact that $pV^{1.2}=c$ means that you can state that $p_1V_1^{1.2}=p_2V_2^{1.2}$, which can be rearranged to give $p_2=\frac{p_1V_1^{1.2}}{V_2^{1.2}}$

ii) $p_2=\frac{800,000*0.03339^{1.2}}{0.09^{1.2}}=243410Pa=0.243MPa$

Finally, rearranging $pV=nRT$ to give $T=\frac{pV}{nR}$ allows us to calculate the final temperature:

iii) $T_2=\frac{243,410*0.09}{4.144*8.31441}=636.4K=363.3\unicode{x2103}$. This can also be verified in the same way as before.

• Thank you for the reply. Can i ask how you found the value of $$'n'$$? As i thought is was $$n = pv/rt$$. But i don't get the same answer as you. And $$T1 = 773.3K$$ is Kelvin. I got 773.15 when i converted it. I had the equations, i just couldn't rearrange them. Feb 7 '18 at 15:35
• "$n$" is the number of moles of gas that you have. This is equal to the mass of gas, divided by the mass of 1 mole of of that gas. For air this is $0.029kg$, so, $\frac{0.12kg}{0.029kg/mol}=4.144mol$. Sorry - I mis-typed my Kelvin conversion the first time - the error didn't carry through, though, and I've edited my answer. Feb 7 '18 at 16:37
• Thank you so much for your help. I just wanted to say that $$\frac {0.12kg}{0.029kg/mol} = 4.137$$ Not $$4.144mol$$ Or am i wrong? Feb 8 '18 at 8:25
• Sorry, I actually used $0.02896kg/mol$ when typing in my calculator, but didn’t show that many decimal places in my working as it would have looked messy - seems I’ve only caused more confusion! wolframalpha.com/input/?i=molar%20mass%20of%20air Feb 8 '18 at 8:58
• Hello, That's fine. I was just wondering. I understand this a lot more now. I have a question i am going to test it on. To see if i can complete it. Thank you for your help. Feb 8 '18 at 10:42

You can also solve the problem without knowing the average molar mass air, which is why R(specific) is given. Multiplying R(specific) with mass will give you the number of moles multiplied by the universal gas constant. i.e mR(specific)=nR Proceed on the usual lines to get the answer after this.

• This is correct. However, if you don't know the average molar mass, and are only given $R_{specific}$, I would recommend working out the average molar mass from the data given using $M=\frac{R}{R_{specific}}$ which allows you to use the more standard $pV=nRT$ formula without confusion. Feb 8 '18 at 17:51