Isothermal Expansion - Rev. or Irre. ?

[zorro]

Homework Statement

Calculate the increase in internal energy of 1 kg of water at 100 degrees Celsius when it is converted into steam at the same temperature and at 1atm. The density of water and steam are 1000kg/m³ and 0.6kg/m³ resp. The latent heat of vaporisation of water is 2.25 x 10⁶ J/kg

The Attempt at a Solution

I am having a difficulty in figuring out whether the isothermal expansion is reversible or irreversible.

Can somebody shed light on it?

 

[Andrew Mason]

Homework Statement

Calculate the increase in internal energy of 1 kg of water at 100 degrees Celsius when it is converted into steam at the same temperature and at 1atm. The density of water and steam are 1000kg/m³ and 0.6kg/m³ resp. The latent heat of vaporisation of water is 2.25 x 10⁶ J/kg

The Attempt at a Solution

I am having a difficulty in figuring out whether the isothermal expansion is reversible or irreversible.

It is bound to be somewhat irreversible but it does not matter. The only thing that matters is that it is isothermal (constant temperature) and isobaric (constant pressure).

Just apply the first law. The trick here is to find the work that is done by the steam in the process of vaporisation.

AM

 

[zorro]

Offcourse it matters, the work done in an irreversible isothermal expansion is different from that in a reversible isothermal expansion.

How do you conclude that it is somewhat irreversible? Can you explain?

 

[RTW69]

You are going from a saturated liquid to a saturated vapor at 100 C and 1 atm. Look on your Properties of saturated water tables at that temperature and pressure. The tables will give you the internal energy for both states. Subtract internal energy for Saturated liquid from internal energy of saturated vapor.

 

[Andrew Mason]

Offcourse it matters, the work done in an irreversible isothermal expansion is different from that in a reversible isothermal expansion.

How do you think it differs?

How do you conclude that it is somewhat irreversible? Can you explain?

A reversible isothermal expansion is a quasi-static expansion that occurs due to an infinitessimal temperature difference with the surroundings (the surroundings being infinitessimally warmer). That is a theoretical limit. In this case, heat is being supplied to change the state. Unless it occurs infinitely slowly, heat is being supplied from a reservoir of some kind that is at a higher temperature than 100C. It does not matter what that temperature is. All that matters is the temperature of the water and steam.

AM

 

[Andrew Mason]

Offcourse it matters, the work done in an irreversible isothermal expansion is different from that in a reversible isothermal expansion.

Why? How is it different?

How do you conclude that it is somewhat irreversible? Can you explain?

Because it does not take an infinitely long time to transfer heat to the water. So there is more than an infinitessimal temperature difference between the hot reservoir and the water.

AM

 

[zorro]

Why? How is it different?

Work done in irreversible I.E. is lesser than that during reversible I.E (in magnitude).

Use W=Pdel(V) for irreversible and W=nRTln(V₂/V₁) for reversible expansion for the above problem and you can verify.

Note that irreversible I.Es are of two types - free and intermediate. In free expansion, work done is 0.

Because it does not take an infinitely long time to transfer heat to the water. So there is more than an infinitessimal temperature difference between the hot reservoir and the water.

AM

I learned something else to differentiate b/w them. If external pressure is less than the pressure inside the system, then it is irreversible. If both pressures are equal, then it is reversible.

Here boiling occurs when the vapour pressure equalises the atmospheric pressure. So they are equal here leading to a reversible process.

 

[Andrew Mason]

Work done in irreversible I.E. is lesser than that during reversible I.E (in magnitude).

Use W=Pdel(V) for irreversible and W=nRTln(V₂/V₁) for reversible expansion for the above problem and you can verify.

Ok. I see where your difficulty is. The issue is not whether the isothermal process is reversible but whether the expansion is against a constant external pressure. It is. If the internal pressure of the steam is greater than external P, the steam expands until it equals external P.

Find the volume dV occupied by the steam molecules created when a heat flow dQ enters the water (non-reversibly). From that determine the expression for dW and integrate to determine total work done.

AM

 

[zorro]

The issue is not whether the isothermal process is reversible but whether the expansion is against a constant external pressure.

How can we conclude from that that the process is irreversible? Is it a rule?

 

[Andrew Mason]

How can we conclude from that that the process is irreversible? Is it a rule?

There is nothing in the problem to indicate that the process is quasi-static. A real process is always irreversible in the real world. But it doesn't matter. You do not have to conclude anything about the reversibility.

You do not need to know how fast the process occurs. All you need to do is apply the first law. The first law applies applies to all real processes.

When an amount of heat dQ vaporises an element of water, dn, the volume it occupies will be (assuming steam is an ideal gas) dV = dnRT/P where P is the external pressure. So the work done will be PdV = dnRT. P is constant throughout. Integrate that over the entire process to determine the total work done (you to also have take into account the initial volume, but it is relatively small). Then apply the first law to find the change in internal energy.

AM

 

[zorro]

I got it now. Thanks!