Applications of Thermodynamics

Question

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 ? — Solar Mike, Feb 07 '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? — Kyle Anderson, Feb 07 '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. — am304, Feb 07 '18 at 14:53
There are far better ways to learn a few equations than watching videos. — Carl Witthoft, Feb 07 '18 at 15:58
For calculating with moles, use $p,V=n,R,T$ with $n$ being the amount of moles and $R\approx 8.314 ,,\mathrm{J/(mol ,K)}$ being the universal gas constant. For calculating with masses, use $p,V=m,R_s,T$ with $m$ being the mass and $R_s$ the specific gas constant ($287,,\mathrm{J/(kg,K)}$ in your case). I personally prefer calculating with masses. Whith $R_s$ being given, I would also calculate with masses. — Robin, Feb 08 '18 at 15:05

Jonathan R Swift · Accepted Answer · 2018-02-07T16:47:36.793

5

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.

edited Feb 07 '18 at 16:47

answered Feb 07 '18 at 15:13

Jonathan R Swift

7,771
1
15
28

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. – Kyle Anderson Feb 07 '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. – Jonathan R Swift Feb 07 '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? – Kyle Anderson Feb 08 '18 at 08:25
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! http://www.wolframalpha.com/input/?i=molar%20mass%20of%20air – Jonathan R Swift Feb 08 '18 at 08:58
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. – Kyle Anderson Feb 08 '18 at 10:42
Hello, I have one more question. Temperature is compressed according to the law $$pV^1.2 = c$$ through ratio 8:1. How do i aplly the ratio the the law $$pV^1.2 = c$$ I have a question i want to try do on my own. But do not know who to apply the 8:1 compression ration. Thanks. – Kyle Anderson Feb 08 '18 at 12:14
If your initial volume is $V_1=100m^3$, then an "8:1 compression ratio" means that your compressed volume is $V_2=\frac{100}{8}=12.25m^3$. – Jonathan R Swift Feb 08 '18 at 13:49
"8:1 compression ratio" means simply that the original volume is 8 times larger than the final volume. It could equally be written as "1:0.125", indicating that the final is 0.125 times the original. Make sense? – Jonathan R Swift Feb 08 '18 at 13:51
Hello, Thank you. I think i understand what you mean. I am just working something out using compression ratio.To see if i can get it right. – Kyle Anderson Feb 08 '18 at 14:27

score 1 · Answer 2 · answered Feb 08 '18 at 17:32

1

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.

answered Feb 08 '18 at 17:32

Divyesh Narayanan

21
2

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. – Jonathan R Swift Feb 08 '18 at 17:51

Applications of Thermodynamics

2 Answers2