A 0.160 mol quantity of nitrogen gas, behaving ideally, and initially at 1.80×10

Question

A 0.160 mol  quantity of nitrogen gas, behaving ideally, and initially at 1.80×105 Pa  and 290 K , undergoes a 3-step process. First, it is compressed isobarically to half its initial volume. Second, it expands adiabatically to its initial volume. Third and finally, it is heated isochorically back to its initial state.

A) What is the heat added to the gas during the compression in the first step?

B) By how much did the internal energy of the gas change during the compression in the first step?

Please show work.

Explanation / Answer

Some important points about N2

1. The value of gama of nitrogen = 1.4

2. The value of Cv = 2.5R

A.) In first process gas is compressed isobarically ( ie. Pressure is kept constant )

Thus work is performed on it W = PV = PVinitial /2 ( since vf = vi /2)

Where P = 1.8×10^5 Pa

And from ideal gas equation Vinitial = nRTi /P = 2.14 × 10^-3 m3

So W = 1.926× 10^2 J

Now change in internal energy of gas

U = n Cv T

Again in isobaric process V is directly proportional to T

So Vf = Vi    imply that Tf = Ti /2 = 145 K

So U = - 0.16 × 2.5 × 8.34 × 145 = - 482.212 J

So heat supplied to gas

H = W + U = - 289.6 J

Again I have already calculated the change in internal energy

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