I have a metal pistonfilled with a room temperature perfect gas at 1 atm of pres

Question

I have a metal pistonfilled with a room temperature perfect gas at 1 atm of pressure.I

compress the gas byapplying a constant ~2 atm. pressure. The gas of course heats up,but

then it cools back toroom temperature.

(a). Is there a changein U and H for this net process? (and why?)

(b). While stillapplying pressure to the compressed piston, I wrap it in insulatingglass

wool. I then suddenlylet go of it. Does the piston expand or compress now that itis

adiabatically cut offfrom the surroundings (ie. no heat can be exchanged)?

(c). If you think thepiston will expand or compress, how do you classify thetransition

and WHY? (the“why” question is worth 100% of the credit)

If you recall, alltransitions are either:

Isothermal,reversible

Isothermal,irreversible

Adiabatic,reversible

Adiabatic, irreversible

Explanation / Answer

a. This question is an application of the first law ofthermodynamics. U = Q + W (Total change in the internal energy) = (Heat added to thesystem) + (Work done by the System) U, the total internal energy will change because the system isnot an isolated one, which means that heat energy can flow in andout of the system. An external pressure is being applied to thesystem, which signifies that work is being done by the system. Alsono heat is being added to added to the system. Therefore, U = 0 -PextV Since U does not equal 0, there is a change in theinternal energy, U. H does not change because at constant pressure H = q Since no energy is added to the system, q = 0; thus, H =0. b. the piston would expand c. The piston would expand because once the constant pressureof 2 atm is removed, the pressure inside the piston is greater thanthe external pressure; thus, it would equalize that pressure byexpanding. The reaction would be considered adibiatic because thereis no heat exchange with the surroundings due to the glass wool. Tobe more specific it would be an adabiatic cooling because thetemperature of the gas would decrease as the pressuredecreases. Since the temperature (not internal energy) decreases,the transition cannot be isothermal. Finally the transition isirreversible because irreversible work occurs when work is doneagainst the external pressure because the internal and externalpressure are not equal. Meaning the pressure inside the pistonis most likely 2 atm, while the external pressure is only 1 atm.Therefore, this transition state is an adibiaticirreversible.

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