Compression and Expansion of a Volume of Air

[Ben_Walker1978]

Homework Statement

Air with an initial volume of 0.12m³, pressure 1 bar and temperature 18°C is compressed according to the law pV1.3 = c through an 8:1 compression ratio. It is then allowed to expand isothermally back to its initial volume. Determine:
i) the pressure and temperature after compression, and

ii) the final pressure.

For air, take R = 287 Jkg¯¹ K¯¹.

Is my solution correct?

Thanks

Homework Equations

The Attempt at a Solution

 

[Chestermiller]

Homework Statement

Air with an initial volume of 0.12m³, pressure 1 bar and temperature 18°C is compressed according to the law pV1.3 = c through an 8:1 compression ratio. It is then allowed to expand isothermally back to its initial volume. Determine:
i) the pressure and temperature after compression, and

ii) the final pressure.

For air, take R = 287 Jkg¯¹ K¯¹.

Is my solution correct?

Thanks

Homework Equations

The Attempt at a Solution

View attachment 220251

1. Compression ratio is defined as the initial volume divided by the final volume, not the final pressure divided by the initial pressure.

2. The pressure-volume equation they gave you was supposed to be $pV^{1.3}=c$, not what you used. This means that the temperature was changing during the compression.

3. Your calculation of the temperature was incorrect, because the volume was not constant. You didn't need to determine the temperature. You need to determine the pressure after the gas is compressed, given the initial volume and pressure, and the final volume. This can be done with the p-V equation they gave you.

 

[Ben_Walker1978]

Hello.

Thank you for your reply.

So if i do 0.12 x 8 = 0.96.

0.96 is final volume.

Becasue 0.96/0.12 = 8. And 8 is Compression Ratio.

Am i correct?

 

[Chestermiller]

No. When something gets compressed, its volume decreases, not increases.

 

[Ben_Walker1978]

So is it 0.12/8?

0.12/8 = 0.015

Final volume = 0.015

 

[Chestermiller]

So is it 0.12/8?

0.12/8 = 0.015

Final volume = 0.015

Correct.

 

[Ben_Walker1978]

Thank you.

So for the 2.

The temperature does not need to change. So if i use $$pV^{1.3}=c$$ to change pressure after compression. How do i show the temperature does not change?

 

[Chestermiller]

You are not done with 1 yet. You need to determine the pressure at the end of 1. You have the equation $$pV^{1.3}=c$$, and you know the initial pressure, the initial volume, and the final volume. Can you figure out how to use this equation to get the final pressure at the end of 1?

 

[Ben_Walker1978]

Final Pressure after compression ratio =
$$ pV^{1.3} = c $$
$$p1V1^{1.3} = p2V2^{1.3}$$
$$p2 = p^1V1^{1.3}$$
$$ V2 =\frac{Initial Volume}{Compression Ratio}$$
$$V2 = \frac{0.12^3}{8}$$
$$p2 = \frac{Inital Pressure x Initail Volume}{Final Volume}$$
$$p2 = \frac{100,000 x 0.12^{1.3}}{0.015^{1.3}}$$
$$ Final Pressure = 423491.1713Pa$$

Is this correct?

 

[Chestermiller]

Final Pressure after compression ratio =
$$ pV^{1.3} = c $$
$$p1V1^{1.3} = p2V2^{1.3}$$
$$p2 = p^1V1^{1.3}$$
$$ V2 =\frac{Initial Volume}{Compression Ratio}$$
$$V2 = \frac{0.12^3}{8}$$
$$p2 = \frac{Inital Pressure x Initail Volume}{Final Volume}$$
$$p2 = \frac{100,000 x 0.12^{1.3}}{0.015^{1.3}}$$
$$ Final Pressure = 423491.1713Pa$$

Is this correct?

No. I have no idea what you did here, but it's not correct.

For step 1, you have $$p_{1f}(V_{1f})^{1.3}=p_{1i}(V_{1i})^{1.3}$$
where $p_{1i}=100000\ Pa$, $V_{1i}=0.12\ m^3$, and $V_{1f}=0.015\ m^3$. So, $$p_{1f}(0.015)^{1.3}=(100000)(0.12)^{1.3}$$
From this equation, what value do you get for the final pressure from step 1, $p_{1f}$?

 

[Ben_Walker1978]

$$p1f$$ = 1492918.214

Is this correct?

 

[Chestermiller]

$$p1f$$ = 1492918.214

Is this correct?

Yes.

 

[Ben_Walker1978]

Thank you.

So that is final pressure after compression ratio.

For temperature after compression ratio. The Temperature doesn't change.

So what equation would i use to solve that? $$pV^{1.3}=c$$ ?

 

[Chestermiller]

Thank you.

So that is final pressure after compression ratio.

For temperature after compression ratio. The Temperature doesn't change.

So what equation would i use to solve that? $$pV^{1.3}=c$$ ?

The final temperature at the end of step 1, the compression step, is not the same as at the beginning of step 1. You can use the ideal gas law to get this temperature.

 

[Ben_Walker1978]

I have used the ideal gas law. This is my attempt.

$$pv = nRT$$

$$T = \frac{pV}{nR}$$

$$T = \frac{1492918.214 x 0.015}{0.5179 x 8.31441} = 359511.5117 ~$$

Is this correct?

Thanks

 

[Chestermiller]

I have used the ideal gas law. This is my attempt.

$$pv = nRT$$

$$T = \frac{pV}{nR}$$

$$T = \frac{1492918.214 x 0.015}{0.5179 x 8.31441} = 359511.5117 ~$$

Is this correct?

Thanks

No. How did you get n = 0.5179 moles? To get the number of moles, use the initial conditions.

Plus, even for the numbers shown on the left hand side of your equation, the number on the right hand side is incorrect. The arithmetic is wrong.

 

[Ben_Walker1978]

I got the number of moles by $$\frac{Mass}{Air}$$

Which is $$\frac{0.015}{0.02896}$$

0.02896 = Air
0.015 is Mass After compression.

Have i got the number of moles wrong?

Yeah it is wrong. It is 52000.59

 

[Chestermiller]

I got the number of moles by $$\frac{Mass}{Air}$$

Which is $$\frac{0.015}{0.02896}$$

0.02896 = Air
0.015 is Mass After compression.

If you had used units, you would know that the 0.015 m^3 is volume, and not mass.

Have i got the number of moles wrong?

Yeah it is wrong. It is 52000.59

I get 4.96 moles.

 

[Ben_Walker1978]

How did you get the mass?

 

[Chestermiller]

How did you get the mass?

From the initial conditions,
$$n=\frac{(100000)(0.12)}{(8.314)(273.2+18)}=4.96$$

 

[Ben_Walker1978]

Thank you.

So this is the equation.

$$ T = \frac{1492918.214 x 0.015}{4.96 x 8.31441} = 543.024 $$

And this is the Temperature after compression.

Correct?

 

[Chestermiller]

Thank you.

So this is the equation.

$$ T = \frac{1492918.214 x 0.015}{4.96 x 8.31441} = 543.024 $$

And this is the Temperature after compression.

Correct?

Yes. And the units of that temperature are ?

 

[Ben_Walker1978]

Kelvin?

 

[Chestermiller]

Kelvin?

You don’t seem sure.

 

[Ben_Walker1978]

well i thought you changed degrees to Kelvin for this equation.

So it's Kelvin

 

[Chestermiller]

well i thought you changed degrees to Kelvin for this equation.

So it's Kelvin

Correct. You can tell that from examining the units of R. So, if this is the final temperature in K, what is this final temperature in C?

OK. Now for step 2. Please say in words you assessment of what's happening here and how to proceed.

 

[Ben_Walker1978]

269.874 Degrees

You have to work out the final pressure.

It is allowed to expand isotheramlly back to its initial volume.

Which means the Temperature is constant.

 

[Chestermiller]

269.874 Degrees

You have to work out the final pressure.

It is allowed to expand isotheramlly back to its initial volume.

Which means the Temperature is constant.

If the temperature is constant, then, from the ideal gas law, the pressure is
(a) directly proportional to the volume
(b) inversely proportional to the volume
(c) directly proportional to the volume to the 1.3 power
(d) inversely proportional to the volume to the 1.3 power
(e) none of the above

 

[Ben_Walker1978]

(b) Inversely proportional to the volume.

 

[Ben_Walker1978]

I have done some revising and found the equation.

$$T = Q - W$$

This is for Temperature constant.

Am i correct?

 

[Chestermiller]

(b) Inversely proportional to the volume.

So, if the pressure is inversely proportional to the volume and you know the initial and final volumes, and the initial pressure, what is the final volume?

 

[Chestermiller]

I have done some revising and found the equation.

$$T = Q - W$$

This is for Temperature constant.

Am i correct?

No.

 

[Ben_Walker1978]

Is the final pressure 800000 Pa.

I done $$P1V1 = P2V2$$

 

[Chestermiller]

Is the final pressure 800000 Pa.

I done $$P1V1 = P2V2$$

This is not correct. You started Step 2 with a pressure of 14.9 bars, and you increased the volume at constant temperature.

 

[Ben_Walker1978]

The volume at constant temperature was 0.12. So is both the volumes 0.12?