In answering this set of nine questions, you are encouraged to draw a PV diagram

Question

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 8.00 liters of an ideal (monatomic) gas at a pressure of 30.0 atm and a temperature of 320 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 =

The temperature of the system is reduced, keeping the volume constant at 8.00 liters, until the pressure in the system equals 9.00. Call this state of the system B. Calculate the temperature at this new state B in degrees K.

Temperature at state B =

Now the gas is allowed to expand at constant pressure (9.00 atm) until the temperature is again equal to 320 K. Call this state of the system C. Calculate the volume of the gas at state C.

Volume at state C =

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 =

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 =

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 =

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 =

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 =

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 =

Explanation / Answer

For the first question, use the ideal gas law to calculate the number of moles of gas in the system:

n = P*V/R*T

n = (30 atm)*(8.00 L)/[(0.0821 L*atm/(mol*K))*(320K)]

n = 9.14 mol

Now use the gas law to calculate the temperature at which the pressure in this amount of gas would be 15.0 atm, assuming the volume remains constant:

T = PV/n*R

T = (15.0 atm)(6.00 L)/[(7.98 mol)*(0.0821 L*atm/(mol*K))]

T = 137 K

Now calculate the volume at 357 K, assuming a pressure of 15 atm:

V = n*R*T/P = (7.98 mol)(0.0821 L*atm/(mol*K))*(357K)/(15 atm)

V = 15.6 L

Strictly speaking, you are not given enough information to calculate the work done by/on the system. You need to know the pressure of the surroundings as the system is changing, not the internal pressure of the system. Assuming that the external pressure (the pressure of the surroundings) is always equal to the pressure inside the system, then the work done by the system on the surroundings is given by:

w = INTEGRAL of {P dV}

If the pressure is constant, then this integral is simply:

w = P*?V

In going from state C to B, the pressure is constant, and the volume changes by -9.6 liters = -9.6*10^-3 m^3, the pressure is 15 atm = 1.52*10^6 Pa, so the work done by the system in this part of the path is:

w_CB = (-9.6*10^-3 m^3)*(1.52*10^6 Pa) = -1.46*10^4 J

In going from state B to state A, there is no volume change, so PdV = 0, and no work is done over this part of the path. The total work done in going from C to B to A is then just w_CBA = -1.46*10^4 J

Note that the question is not clear enough to define the sign convention. I've defined work as the work done *by* the system, therefore the negative sign indicates that the system does negative work, i.e., positive work is done on the system by the surroundings. Alternatively, if one defines the work as the work done on the system by the surroundings, then the work would be positive.

The heat absorbed by the system in going from state C to B is given by:

q_CB = C*n*(T_B - T_C), where C is the molar heat capacity, n is the number of moles, and T_B and T_C are the temperatures of states B and C.

q_CB = (5/2)*(0.0821 L*atm/(mol*K))*(137K - 357K) = -3.65*10^4 J

The absorbed heat is negative, indicating that the system has lost thermal energy in this step.

Because this system is an ideal gas, and the temperatures of states A and C are the same, the internal energy of states A and C are the same. (The internal energy of an ideal gas depends only on temperature.) From the first law of thermodynamics, we know that:

?E = q - w

where w is the work done by the system on the surroundings and q is the heat added to the system.

In terms of what we;ve already calculated, we have that:

?E = 0 = q_CB + q_BA - w_CBA

0 = -3.65*10^4 J + q_BA - (-1.46*10^4 J)

Now solve for q_BA, the heat added to the system along the path from B to A:

q_BA = 2.19*10^4 J

just change the numbers

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