# 1 dm^3 of water steam at 200 degrees C, 10 bar enters a compressor. When it leaves, its temperature is 400 deg. C and volume 0.5 dm^3. Calculate the final pressure of steam. Critically discuss the assumptions you made in your calculations.

• 770 views

This question refers to the behavior of gases under compression. In IB we model such situations using the ideal gas equation, which states that:

pV = nRT,

where:

p - pressure of the gas

V – its volume

n – number of moles of the gas

R – gas constant

T – temperature of the gas

We are given a description of the gas at two different states, so begin by writing down the equations for state 1 and state 2

p1V1 = n1R1T1

p2V2 = n2R2T2

Now notice that the amount of gas entering the compressor and leaving it should be the same, so n1 = n2 = n and of course the gas constant is the same in both cases (let’s just call it R).

So rewriting the equations again:

p1V1 = nRT1

p2V2 = nRT2

Rearranging them both to receive the expression for n”

n = (p1V1)/(RT1)

n = (p2V2)/(RT2)

Since this is the same n we can say that

(p1V1)/(RT1) =(p2V2)/(RT2)

The R is on both sides of the equation so we can cross it out:

(p1V1)/(T1) =(p2V2)/(T2)

We are asked to calculate p2, so rearrange this equation to obtain an expression for it:

p2 = (p1V1T2)/(T1V2)

So now we can just substitute the relevant values and get the final result, right? Well, not really, because if you have a closer look at them you will notice that they are not in SI units. We need to convert:

The temperatures to Kelvin:

T1 = 200 C = 473 K

T2 = 400 C = 673 K

Volumes to m^3:

V1 = 1 dm^3 = 0.001 m^3

V2 = 0.5 dm^3 = 0.0005 m^3

Pressures to Pa:

p1 = 10 Bar = 1000000 Pa

Now we can just plug then into the equation to get:

p2 = 2850000 Pa = 2.85 MPa

As you can see, both pressures are quite high (in the range of MPa). In our calculations we assumed that steam will behave like an ideal gas, i.e there will be no interactions between its molecules. At such pressures they will be so close together (remember your kinetic gas theory?), that they might indeed interact, so the results we got may not be a very good model of this situation. In fact, if you made measurements of pressure in states 1 and 2, you will find out that your results are off by approximately 10%.

Still stuck? Get one-to-one help from a personally interviewed subject specialist.

95% of our customers rate us

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this.