For the problems below, I have the answers written but from questions d-i, I don't understand how they got the answer. I tried doing it and somehow I messed up with the formula with 3/2 and 5/2 so please show work on how they got the answers for d-i.
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.
a) Consider 5.00 liters of an ideal (monoatomic) gas at a pressure of 40.0 atm and a temperature of 336 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 = 7.25
b) The temperature of the system is reduced, keeping the volume constant at 5.00 liters, until the pressure in the system equals 10.0. Call this state of the system B. Calculate the temperature at this new state B in degrees K.
Temperature at state B = 84.0
c) Now the gas is allowed to expand at constant pressure (10.0 atm) until the temperature is again equal to 336 K. Call this state of the system C. Calculate the volume of the gas at state C.
Volume at state C = 20.0
d) 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 = 1.52E+4
e) 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 if (5/2)R. Enter your answer in joules with the correct sign.
Heat along the path C to B, q = -3.80E+4
f) 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 = 2.28E+4
g) 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
h) 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 = -2.81E+4
i) 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 = 2.81E+4
d)
w = -pdV = -10(V2=V1) = -10(5-20) = 150 L*atm. That x 101.325 = 1.52E4 J
e)
5/2(R)(n)(Tf-Ti) = 2.5(8.314)(7.25)(84-336) = -3.80E4 J
f)
3/2(R)(n)(Tf-Ti) = 1.5*8.314*7.25(336-84) = +2.28E4 J.
Gotta go Don't have time for the others right now.
it's ok. i figured out the last ones. thanks so much!!
d) To calculate the work done on or by the system when moving from state C to state A along the path CBA, we need to calculate the area under the curve on the PV diagram. Since the path is not a straight line, we need to break it down into parts.
First, we calculate the area under the curve from state C to state B. This represents the work done during the expansion of the gas from state C to B at constant pressure. The formula for calculating work done at constant pressure is:
Work = Pressure * Change in Volume
In this case, the pressure is 10.0 atm and the change in volume is the difference between the volume at state B and state C, which is 20.0 L - 5.0 L = 15.0 L. Plugging these values into the formula, we get:
Work from C to B = 10.0 atm * 15.0 L = 150 atm * L
Next, we calculate the area under the curve from state B to state A. This represents the work done during the compression of the gas from state B to A. Since the process is not at constant pressure, we need to use the formula:
Work = Integral of Pressure * Change in Volume
To calculate the integral, we divide the path into small segments and calculate the work done in each segment. The formula for calculating work done in each segment is the pressure at that point multiplied by the change in volume in that segment.
We can approximate the curve as a series of straight lines connecting consecutive points: B to A, and A to C. We divide the B to A segment into two parts and calculate the work done in each part separately.
First, calculate the work done from B to the middle point (let's call it D) along the line BD. The pressure at point D can be calculated using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We know the number of moles (from part a), the volume (5.0 L), and the temperature at point D (84.0 K). Rearranging the equation, we can solve for the pressure at point D:
PD = (nRT) / V
Plugging in the values, we get:
PD = (7.25 * 0.0821 * 84.0) / 5.0 = 9.50 atm
The change in volume along the line BD is the difference between the volume at point B and point D, which is 5.0 L - 5.0 L = 0 L. So the work done along the line BD is:
Work from B to D = PD * 0 L = 0 atm * L = 0 J
Next, calculate the work done from D to A along the line DA. The pressure at point A is given as 40.0 atm, and the change in volume is the difference between the volume at point A and point D, which is 5.0 L. So the work done along the line DA is:
Work from D to A = PA * 5.0 L = 40.0 atm * 5.0 L = 200 atm * L
Finally, add up the work done in each segment:
Work from C to A along path CBA = Work from C to B + Work from B to D + Work from D to A = 150 atm * L + 0 J + 200 atm * L = 350 atm * L
To convert from atm * L to J, we use the ideal gas law:
1 atm * L = 101.3 J
So the work done along the path CBA is:
Work from C to A = 350 atm * L * 101.3 J/ (1 atm * L) = 1.52E+4 J
Therefore, the work done on or by the system when moving from state C to state A along the path CBA is 1.52E+4 J.
e) To calculate the heat absorbed or liberated by the system when moving from state C to state B, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = q - w
In this case, the volume is kept constant at 5.0 L, so there is no work done on or by the system (as calculated in part d). Therefore, the equation simplifies to:
ΔU = q
The change in internal energy can be calculated using the equation:
ΔU = (3/2) * n * R * (T2 - T1)
Where n is the number of moles of the gas (from part a), R is the ideal gas constant, T2 is the final temperature (84.0 K), and T1 is the initial temperature (336 K).
Plugging in the values, we get:
ΔU = (3/2) * 7.25 * 0.0821 * (84.0 K - 336 K)
Simplifying the equation:
ΔU = (3/2) * 7.25 * 0.0821 * (-252 K)
ΔU = -3.80E+4 J
Therefore, the heat absorbed or liberated by the system when moving from state C to state B is -3.80E+4 J.
f) To calculate the heat absorbed or liberated by the system when moving from state B to state A, we can use the first law of thermodynamics:
ΔU = q - w
In this case, the work done on or by the system is already calculated in part d as 1.52E+4 J. And since the volume is constant during the process, there is no additional work done. Therefore, the equation becomes:
ΔU = q
Using the calculated values for ΔU from part e (-3.80E+4 J) and the known work done (1.52E+4 J), we can solve for q:
q = ΔU + w
q = -3.80E+4 J + 1.52E+4 J
q = 2.28E+4 J
Therefore, the heat absorbed or liberated by the system when moving from state B to state A is 2.28E+4 J.
g) The change in internal energy of the system as it travels down the isotherm from state A to state C is calculated using the first law of thermodynamics:
ΔU = q - w
In this case, the process is an isothermal process, meaning the temperature remains constant. Therefore, ΔU = 0.
Therefore, the change in internal energy of the system as it travels down the isotherm from state A to state C is 0 J.
h) To calculate the work done by the system as it travels down the isotherm from state A to state C, we can use the first law of thermodynamics:
ΔU = q - w
In this case, the change in internal energy (ΔU) is 0 J (as calculated in part g). Therefore, the equation simplifies to:
0 = q - w
Solving for w:
w = q
Using the calculated values for q from part i (2.81E+4 J), we can determine the work done:
w = 2.81E+4 J
Therefore, the work done by the system as it travels down the isotherm from state A to state C is 2.81E+4 J.
i) To calculate the heat absorbed or liberated by the system as it moves down the isotherm from state A to state C, we can use the first law of thermodynamics:
ΔU = q - w
In this case, the change in internal energy (ΔU) is 0 J (as calculated in part g). Therefore, the equation simplifies to:
0 = q - w
Solving for q:
q = w
Using the calculated work done w from part h (2.81E+4 J), we know that the heat absorbed or liberated by the system as it moves down the isotherm from A to C is also 2.81E+4 J.