A sample consists of one mole of Ideal Gas molecules. The specific heat at const

Question

A sample consists of one mole of Ideal Gas molecules. The specific heat at constant volume is 16.8 J/mK. The gas is initially at 1 atm and 310K. It undergoes reversible adiabatic expansion until its pressure reaches 5 atm. a) Is it a mono-atomic gas (like Ar), a gas of linear molecules (like 02), or a gas of non-linear molecules (like CH4)? (5 pts) b) Give an argument based on the First Law of thermodynamics why the temperature should drop during adiabatic expansion. (5 pts) c) Calculate the final temperature, the final volume, and the work done by the surroundings. (15 pts)

Explanation / Answer

We know,

Cp-Cv = R

Given : Cv = 16.8 J/(mol.K)

R = 8.314 J/(mol.K)

Thus, Cp = Cv + R = 25.11 J/(mol.K)

Thus, specific heat ratio, r = Cp/Cv = 25.11/16.8 = 1.50

Also, r = 1 + 2/f, where f = degrees of freedom of molecule

For a monoatomic gas,f = 3, so r = 1.67, so it is not a monoatomic gas

It is probably a linear diatomic gas

(2)

During an adiabatic expansion, no heat is supplied to the gas

From first law, we have :

Q = dE + W

Thus, W = -dE

Since in expansion gas needs energy, this energy is derived from its own internal energy itself. Thus internal energy falls and thus temperature falls ( because temperature is directly related to internal energy )

(3)

In this case, since initial pressure is 1 atm and final is 5 atm, this means that gas is being compressed.Thus there is something wrong with the problem statement, because the gas cannot expand with an increase in pressure in an adiabatic process. I am solving this part assuming that the gas is compressed ( and not expanding )

For an adiabatic process, we have :

P11-rT1r = P21-rT2r

Putting values, we get :

11-1.53101.5 = 51-1.5T21.5

Thus, T2 = 530.08 K

P2 = 5 atm = 506625 Pa

n = 1

P2V2 = nRT2

Thus, V2 = 8.69*10-3 m3

Change in internal energy, dE = n*Cv*dT = 1*16.8*220.08 = 3697.344 J

Since q = 0

So, work done by the gas = -3697.344 J

Thus, work done by surroundings = +3697.344 J

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