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 $R_{specific} = 287 Jkg^{-1} K^{-1}.$

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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.

Answer 1

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.

Answer 2

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.