CHEM1101 Midterm: State Functions

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30 May 2018

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State Functions

• Usually e hae o ay of koig the iteral eergy of a syste; fidig that alue is

simply too complex a problem.

• Hoeer, e do ko that the iteral eergy of a syste is idepedet of the path y

which the system

achieved that state.

– In the system

below, the water

could have reached

room temperature

from either

direction.

Therefore, internal energy is a state function.

• It depeds oly o the preset state of the syste, ot o the

path by which the system arrived at that state.

• Ad so, DE depeds oly o Eiitial ad Efial.

However, q and w are not state functions.

• Whether the attery is shorted out or is disharged y ruig

the fan, its DE is the same.

– But q and w are different in the two cases

Work

Usually in an open container the only work done is by a gas pushing on the surroundings (or

by the surroundings pushing on the gas).

We can measure the work done by the gas if the reaction is done in a vessel that has been

fitted with a piston (see below image for piston set up):

 = −PDV

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Document Summary

In the system below, the water could have reached room temperature from either direction. Therefore, internal energy is a state function: it depe(cid:374)ds o(cid:374)ly o(cid:374) the prese(cid:374)t state of the syste(cid:373), (cid:374)ot o(cid:374) the path by which the system arrived at that state, a(cid:374)d so, de depe(cid:374)ds o(cid:374)ly o(cid:374) ei(cid:374)itial a(cid:374)d efi(cid:374)al. However, q and w are not state functions: whether the (cid:271)attery is shorted out or is dis(cid:272)harged (cid:271)y ru(cid:374)(cid:374)i(cid:374)g the fan, its de is the same. But q and w are different in the two cases. Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas). We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston (see below image for piston set up): (cid:449) = pdv.

Related Questions

In answering this set of nine questions, you are encouraged to draw a PV diagram, with P (pressure) on the y axis and V (volume) on the x axis. Plot the three points A, B, and C on this diagram.

Consider 3.00 liters of an ideal (monatomic) gas at a pressure of 31.0 atm and a temperature of 356 K. Call this state of the system A. Using the ideal gas law, calculate the number of moles of gas present in the system.

Number of moles, n = 3.19

The temperature of the system is reduced, keeping the volume constant at 3.00 liters, until the pressure in the system equals 16.0. Call this state of the system B. Calculate the temperature at this new state B in degrees K.

Temperature at state B = 184

Now the gas is allowed to expand at constant pressure (16.0 atm) until the temperature is again equal to 356 K. Call this state of the system C. Calculate the volume of the gas at state C.

Volume at state C = 5.81

Now calculate the work done on or by the system when the system moves back from state C to state A along the path CBA. Enter your answer in joules with the correct sign.

Work along the path CBA, w = 4.56E+3

Calculate the heat absorbed or liberated by the system when the system moves from state C to state B. Note that the molar heat capacity of an ideal gas under constant pressure is (5/2)R. Enter your answer in joules with the correct sign.

Heat along the path C to B, q = -1.14E+4

Using your results from the previous two questions, calculate the heat absorbed or liberated by the system as it moves from state B to state A. Enter your answer in joules with the correct sign.

Heat along the path B to A, q = 6.84E+3

What is the change in the internal energy of the system as the system travels down the isotherm from state A to state C?

Change in internal energy down the isotherm = 0.000

How much work is done by the system as the system travels down the isotherm from state A to state C? Enter your answer in joules with the correct sign.

Work down the isotherm, w = -6.23E+3

How much heat must be absorbed or liberated by the system as it moves down the isotherm from A to C? Enter your answer in joules with the correct sign.

Heat down the isotherm, q = 6.23E+3

An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 50.0 atm and releases 60.4 kJ of heat. Before the reaction, the volume of the system was 7.80 L . After the reaction, the volume of the system was 2.80 L . Calculate the total internal energy change, ÎE, in kilojoules. Express your answer with the appropriate units. ÎE = -35.2 kJ SubmitHintsMy AnswersGive UpReview Part Correct Although the system absorbed some energy in the form of work, the significant release of heat caused the total change to be negative. State functions versus path functions The change in internal energy, ÎE, is a state function because it depends only on the initial and final states of the system, and not on the path of change. In contrast, q and w are path functions because they depend on the path of change and not just the initial and final states of the system.

Part C An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) confined to a container with a massless piston at the top. (Figure 2) A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 6.60 to 3.30 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 3.30 to 2.64 L . In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 6.60 to 2.64 L in one step. If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules? Express your answer with the appropriate units.

please answer for Part C

CHEM1101 Midterm: State Functions

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