How Much Work Is Done in Adiabatic Compression of Gas?

[v_pino]

Homework Statement

When a volume 10^-3 m^3 of a certain gas at a pressure of 1 atm undergoes a reversible adiabatic compression to half its volume, its pressure rises by a factor of three. The internal energy of the gas is given by E=3PV, where P is the pressure and V is the volume. By making use of the first law of thermodynamics, or otherwise, calculate how much work is done on the gas to perform the compression.

Answer: 152 J

Homework Equations

(1) I tried writing E=E(V,T) or E=E(P,T).

(2) First law of thermodynamics: dE=dQ+dW

The Attempt at a Solution

dQ = 0 for reversible process.
I differentiated (1) but had no way of getting T.

Am I on the right track?

 

[G01]

Homework Statement

When a volume 10^-3 m^3 of a certain gas at a pressure of 1 atm undergoes a reversible adiabatic compression to half its volume, its pressure rises by a factor of three. The internal energy of the gas is given by E=3PV, where P is the pressure and V is the volume. By making use of the first law of thermodynamics, or otherwise, calculate how much work is done on the gas to perform the compression.

Answer: 152 J

Homework Equations

(1) I tried writing E=E(V,T) or E=E(P,T).

(2) First law of thermodynamics: dE=dQ+dW

The Attempt at a Solution

dQ = 0 for reversible process.
I differentiated (1) but had no way of getting T.

Am I on the right track?

Yes. You know that there was no heat input into the gas, because the process is adiabatic. Therefore the change in internal energy of the gas is entirely due to work done on the gas:

[tex]dE=dW[/tex]

Applying this statement, along with the equation given in the problem should be enough to get you to the answer.

 

[v_pino]

I'm getting 150J instead of 152J as given in the answer. I substituted V1 and P1 into equation of E to get E1 and substituted V2 and P2 into equation of E to get E2.

E1 = 300 and E2 = 450.

Subtracting E1 from E2 gives 150J.

 

[G01]

I'm getting 150J instead of 152J as given in the answer. I substituted V1 and P1 into equation of E to get E1 and substituted V2 and P2 into equation of E to get E2.

E1 = 300 and E2 = 450.

Subtracting E1 from E2 gives 150J.

That's fine. You're doing the problem correctly. You're just rounding. Thus, the answer you get is rounded. I compute, using the same method:

E1=303.98J E2=455.96J Thus, W~152J

 

[rwisz]

Your attempt at a solution is on the right track. To find the work done on the gas during the compression, you can use the first law of thermodynamics, which states that the change in internal energy (dE) of a system is equal to the heat added (dQ) to the system minus the work done (dW) by the system. In this case, since the process is adiabatic, there is no heat added (dQ=0). Therefore, the work done on the gas is equal to the change in internal energy (dE).

To find the change in internal energy, you can use the given equation E=3PV and substitute in the values for the initial and final pressures and volumes. The initial pressure is 1 atm and the initial volume is 10^-3 m^3. The final volume is half of the initial volume, or 5x10^-4 m^3. The final pressure is given as three times the initial pressure, so it is 3 atm.

Substituting these values into the equation, we get:
dE=(3x3 atm)(5x10^-4 m^3)-(3x1 atm)(10^-3 m^3) = 152 J

Therefore, the work done on the gas during the compression is 152 J.