Find ΔU , ΔH , q and w for this change in state.

Question

A 16.0 LL volume of an ideal gas in a cylinder with a piston is at a pressure of 3.0 bar . Enough weight is suddenly removed from the piston to lower the external pressure to 1.5 bar . The gas then expands at constant temperature until its pressure is 1.5 bar. Find ΔU , ΔH , q and w for this change in state.

solution;

i got the solution for ΔU as 0 which was right. I thought the value for ΔH was also 0 due to the below reason.

When the temperature of the system is zero or constant then the change in the enthalpy will be equal to zero hence, change in the internal energy  and change in the enthalpy  = 0.

But the correct answer was ΔH= 2400J. May I know why? Also I am unable to find the q and w.

 Highly apprecitate your help. Thanks

Answer 1

So we have a change in pressure at constant temperature. With the ideal gas law, we can find the new volume:

P1V1=P2V2

(3.0 bar)(16.0L) = (1.5 bar)V2. V2 = 32L. So the change in volume is 32-16=16L.

The change in pressure is 3.0 bar - 1.5 bar = 1.5 bar.

To find ΔH, we use ΔH = ΔU + PV. You got it that ΔU = 0.

Plugging in the ΔU, we have ΔH = 0 + PV. So ΔH = PV.

Before we calculate for PV, let's make sure the units match up. We want the final answer in J, so let's work backwards:

J = Pa•m³

1 bar = 10⁵ Pa, 1.5 bar = 1.5•10⁵ Pa

1 L = 0.001 m³, 16 L = 0.016 m³

Now, we can do the final calculations.

ΔH = PV = (1.5•10⁵ Pa)(0.016 m³) = 2400 Pa•m³ = 2400 J

Now for q and w, we use ΔU = q + w and w = -PV. Plugging in ΔU and rearranging, we have q = -w.

Let's calculate for w first:

w = -PV = -2400 J

A quick rule of thumb to check for w is that in gas expansion, w should be negative. In gas compression, w is positive.

Then for q we just do:

q = -w = 2400 J.

There you go.