Thermodynamics me with this problem

[psan6]

a piston cylinder device contain 11 pounds of nitrogen gas at 58 psia and 90 degree farenheit. during the process the gas follows PV=C expansion which 15 BTU of work is done by the system, determine:
a) charge of internal energy (BTU/lb)
b)charge in entholpy (KJ)
c) heat transferred in BTU
d) final pressure in bar units

 

[psan6]

how could i find the internal energy if it has only 1 temperature given??how could i know Cv??thanks

 

[Count Iblis]

PV = C implies that temperature is constant, because you have in general that PV = N k T.

Constant T, in turn implies for an ideal gas that the internal energy is constant, http://en.wikipedia.org/wiki/Intern...change_in_temperature_and_volume_or_pressure"

The internal energy change is also equal to heat absorbed by the gas minus the work performed by the gas. So, because the interna kenergy change is zero, we have:

Work done= heat absorbed = 15 BTU

And the entropy change is absorbed heat/(absolute temperature)

 

[psan6]

if that's the case.why is it internal energy should be in BTU/lb??

 

[Count Iblis]

if that's the case.why is it internal energy should be in BTU/lb??

Well, the answer is 0 BTU/lb

Enthalpy change is also zero, because H = E + P V and P V stays constant.

 

[psan6]

how about the final pressure?its also the same 58 psia?

 

[Count Iblis]

how about the final pressure?its also the same 58 psia?

Work done equals the integral of P dV. If P V = C, that means that P = C/V and the work done equals C Log(V2/V1). Equate this to 15 BTU, compute the constant C from the ideal gas law (PV = N k T), and then find the ratio V2/V1. It then follows from P V = C, that:

P2/P1 = V1/V2

So, you know by what factor the pressure changes. The initial pressure is known...

 

[psan6]

i get this equation 15/C=in(V2/V1), how can i get C constang in PV=NkT?this is what i did..

(58psia)V=N(1.38066X10-23 J/K)(90F)

how can i get N?and V?

and if i get V, what is it V1 or V2?

thanks for answering my question..

 

[Count Iblis]

The temperature needs to be expressed in Kelvin. N is the number of molecules. You have 11 pounds of nirogen gas, the mass of one nitrogen molecule is 28 atomic mass units. One atomic mass unit is 1.66054*10^(-27) kg

You don't need to compute the initial volume. You just compute
N k T. This is your C. Pressure times volume will be equal to this C.

 

[psan6]

then that equation will be like this V2/V1=e^15/c ?? right? how can i convert 11 punds of nitrogen gas into molecules? i can't undserstand it..sorry for that,..thanks again

 

[psan6]

can i know the N of 11 pounds of nitrogen?because i can't get it...

 

[hdsncts]

PV = C implies that temperature is constant, because you have in general that PV = N k T.

Constant T, in turn implies for an ideal gas that the internal energy is constant, http://en.wikipedia.org/wiki/Intern...change_in_temperature_and_volume_or_pressure"

The internal energy change is also equal to heat absorbed by the gas minus the work performed by the gas. So, because the interna kenergy change is zero, we have:

Work done= heat absorbed = 15 BTU

And the entropy change is absorbed heat/(absolute temperature)

I disagree. dU = TdS - PdV, so a change in volume definitely accounts for a change in internal energy.

 

[hdsncts]

PV work (expansion of a piston) takes from the internal energy of a system.

 

[Count Iblis]

PV work (expansion of a piston) takes from the internal energy of a system.

If the temperature is kept constant, then dU = 0 and thus
T dS (= supplied heat) = P dV (= work done)

 

[hdsncts]

If the temperature is kept constant, then dU = 0 and thus
T dS (= supplied heat) = P dV (= work done)

Why do you keep saying that dU = 0 if dT is zero? Please refer to my equation:

dU = TdS - PdV

If either T and dS are nonzero, or P and dV are nonzero, then there is a change in internal energy (dU =/= 0).

T is always going to be nonzero (especially in this case). dS will be greater than zero because as the piston expands in volume, there is an increase in entropy. The pressure may or may not be nonzero at the end, and dV is definitely positive because the piston is expanding.

I still think that dU cannot be zero in this case, although I may be mistaken greatly. In order to explain your opinion please either use the equation above, or tell me why it is not applicable.

 

[Count Iblis]

http://en.wikipedia.org/wiki/Intern...change_in_temperature_and_volume_or_pressure". This section of the wiki page was written by me, b.t.w.

Anyway, to summarize, in this problem it is given that PV = C, meaning that the temperature is kept constant. Then the amount of work done by the gas is given. The heat supplied to the gas is not given, but that follows from the given temperature change (which is zero). I claim that dT = 0 implies that dU = 0, so that the supplied heat to the gas must equal the work done by the gas. That then makes the rest of the problem trivial.

So, how do we know that dU = 0? The general equation is:

dU = T dS - P dV

It is convenient to rewrite this in terms of dT and dV. So, we want to express dS in terms of dT and dV. If you follow the derivation given in the wiki article, you see that dU can be written in terms of dT and dV as:

$$dU =C_{V}dT +\left[T\left(\frac{\partial P}{\partial T}\right)_{V} - P\right]dV$$

If you now use that we are dealing with an ideal gas, so that

$$\left(\frac{\partial P}{\partial T}\right)_{V} =\frac{P}{T}$$

you get:

$$dU = C_{V}dT$$

This means that the internal energy does not depend on the volume if we keep the temperature constant. This, in fact, implies that U considered as a function on T and any other variable X independent of T, will always be independent of X.

The heat capacity is also only a function of the temperature for an ideal gas. This follows from the fact that is is the derivative of U w.r.t. T at constant V. If you differentiate it w.r.t. sme vatriable X independent of T at constant T, and then in this second derivative interchange the order of differentiation and use that the inner derivative is zero, you get the desired result.