University College - Chemistry Chem 112A Lecture Notes - Lecture 6: Enthalpy, Jmol, Isothermal Process

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salmonyak403

27 Feb 2016

School

WUSTL

Department

University College - Chemistry

Course

University College - Chemistry Chem 112A

Professor

Mabbs Richard

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Document Summary

Energy can neither be created nor destroyed, only converted from one form to another. Energy added to a specific region as heat, q. Energy added to a specific region by working on it, w. Internal energy of an ideal monoatomic gas u = n*3/2rt. Work, w: amount of energy added to something by working on it (negative value implies energy loss from something by it working). Work (w) = f ( force) * d (distance) External force (f) = pressure (external) * a (area) w = -area = - pressure * change in volume units w in j volume = m^3. & since change in u = q + w q = -w expansion stops when p_sys = p_ext isothermal system of an ideal gas. Reversible process (isothermal) process which can be reversed by an infinitesimally small change of a variable. State 1: psys = 10atm, t = 298k, v = 2445l n = 1 mole.

Related Questions

One mole of an ideal monatomic gas is inititally at 300 K and 1010 J/L pressure inside a cylinder with a frictionless piston. the cylinder is an isolated (adiabatic) system. the gas expands against zero external pressure. when the volume expands to 24.7 L, a peg stopes the piston. calculate deltaU (the change in energy), q (the heat transferred to the system), and w (the work done by the system) when the expansion is isothermal and reversible.

Energy cannot be created nor destroyed, but it can be transferred between a reaction and its surroundings. The change in energy, Delta E, is positive if the reaction absorbs energy, and it is negative if the reaction releases energy. You may also see this expressed in terms of change in internal energy, Delta U. For the purposes of this question, Delta U and Delta E are equal. One way a reaction can transfer energy to or from the surroundings is by releasing or absorbing heat. A reaction can also transfer energy, in the form of work, through a change in volume. The total change in energy is the sum of the heat and work: Delta E = q + w At constant pressure, q = Delta H and w = -P Delta V and so Delta E = Delta H - P Delta V Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system, based on the following descriptions: Surroundings get colder and the system decreases in volume. Surroundings get hotter and the system expands in volume. Surroundings get hotter and the system decreases in volume. Surroundings get hotter and the system does not change in volume. Also assume that the magnitude of the volume and temperature changes are similar among the reactions. Rank from most energy released to most energy absorbed. To rank items as equivalent, overlap them. Energy cannot be created nor destroyed, but it can be transferred between a reaction and its surroundings. The change in energy, Delta E, is positive if the reaction absorbs energy, and it is negative if the reaction releases energy. You may also see this expressed in terms of change in internal energy, Delta U. For the purposes of this question, Delta U and Delta E are equal. One way a reaction can transfer energy to or from the surroundings is by releasing or absorbing heat. A reaction can also transfer energy, in the form of work, through a change in volume. The total change in energy is the sum of the heat and work: Delta E = q + w At constant pressure, q = Delta H and w = -P Delta V and so Delta E = Delta H - P Delta V A mole of X reacts at a constant pressure of 43.0 atm via the reaction X(g) + 4Y(g) - 2Z(g), Delta H degree = -75.0 kJ Before the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume was 2.00 L Calculate the value of the total energy change. AE, in kilojoules. Express your answer with the appropriate units.

This problemdemonstrates that heat and work are dependent uponpath whileinternal energy is a state function (path independent).Considerthe isothermal expansion of one mole of an ideal gas fromaninitial state (Pinit =1atm and Tinit= 300 K) to a finalstate(Pfinal=0.1 atm, Tfinal=300K). Now this can be
accomplished by an infinite number of paths â lets considerjusttwo.

Path #1 The irreversible isothermal expansion takes place againstaconstant external pressure of 0.1atm.
Path#2 The isothermal expansion takes place in two steps, firstanirreversible expansion to the intermediate state (Pint=0.5atm,Tint=300K) against an external pressure of 0.5 atm, and nextanirreversible expansion from the intermediate state to thefinalstate against an external pressure of 0.1
atm.

a) What is the change in internal energy(ÎU=UfinalâUinit) for path#1? for path #2?
b) What is the work done by the system in path#1? Use the firstlawto calculate the heat absorbed by the system for path#1.
c) What is the work done by the system in path#2? (Sum up theworkfor each step). Use the first law to calculate the heatabsorbed bythe system for path#2.
d) You should have observed that the heat and work weredifferentfor each path although the change in internal energywasindependent of path. Suggest a third path which takes thesystemfrom the same initial state to the same final state.

University College - Chemistry Chem 112A Lecture Notes - Lecture 6: Enthalpy, Jmol, Isothermal Process

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