Consider a system consisting of 2moles CO2 gas initially at 25 deg and 10 atm and confined to a cylinder of cross section 10.0 square centimeter. lt is allowed to expand adiabatically against an external pressure of 1.0 atm until the piston has moved outwardsthrough 20cm. Assume that CO2 may be considered a perfect gas with Cv,m 28.8. Calculate q, w, delta U, delta T and delta S.
To calculate the values you asked for, we need to follow a few steps:
Step 1: Find the final volume of the gas.
The initial volume of the gas is not given, but we are given the cross-sectional area of the cylinder (10.0 square centimeters) and the distance moved by the piston (20 cm). Therefore, we can calculate the initial volume (Vi) using the formula Vi = (cross-sectional area) * (distance moved by the piston).
Vi = (10.0 cm^2) * (20 cm) = 200 cm^3
Step 2: Calculate the final volume (Vf) of the gas.
Since the process is adiabatic, the equation PV^(gamma) = constant can be used, where P is the pressure, V is the volume, and gamma is the ratio of specific heat capacities (Cp/Cv). For CO2, gamma is approximately 1.4.
Using the initial and final pressures, P1V1^(gamma) = P2V2^(gamma), we can find Vf:
(10 atm)(200 cm^3)^(1.4) = (1 atm)(Vf)^(1.4)
Vf = (10 atm)(200 cm^3)^(1.4) / (1 atm)^(1.4)
Step 3: Calculate the work done by the gas (w).
Since the process is adiabatic, we can use the formula: w = (P1V1 - P2V2) / (gamma - 1).
w = (10 atm)(200 cm^3 - Vf) / (1.4 - 1)
Step 4: Calculate the change in internal energy (delta U).
Since the process is adiabatic, there is no heat exchange (q = 0). Therefore, delta U = w.
delta U = w
Step 5: Calculate the change in temperature (delta T).
The equation for adiabatic processes is: P1V1^(gamma) = P2V2^(gamma), which can be rearranged to find the relationship between temperature and volume: T1V1^(gamma-1) = T2V2^(gamma-1).
T1V1^(gamma-1) = T2V2^(gamma-1)
T2 = T1(V1/V2)^(gamma-1)
Step 6: Calculate the change in entropy (delta S).
The change in entropy can be calculated using the formula: delta S = Cv * ln(T2/T1) + R * ln(V2/V1), where Cv is the molar heat capacity at constant volume and R is the gas constant.
delta S = (28.8 J/mol*K) * ln(T2/T1) + (8.314 J/mol*K) * ln(V2/V1)
Now you can plug in the values you have obtained in the previous steps to calculate q, w, delta U, delta T, and delta S.
To solve this problem, we need to use the first law of thermodynamics and the ideal gas equation.
1. Find the initial volume:
Using the ideal gas equation: PV = nRT
We know the pressure (P = 10 atm), the number of moles (n = 2 moles), and the temperature (T = 25°C = 298 K). Rearranging the equation, we can solve for volume (V1):
V1 = nRT / P
Substituting the values, we get:
V1 = (2 moles)(0.0821 L.atm/mol.K)(298 K) / (10 atm)
V1 = 49.16 L
2. Find the final volume:
We are given that the piston has moved outward by 20 cm. Since the cross-sectional area of the cylinder is 10.0 square centimeters, the change in volume (ΔV) is calculated as:
ΔV = (20 cm)(10.0 cm^2)
ΔV = 200 cm^3 = 0.2 L
The final volume (V2) is the sum of the initial volume and the change in volume:
V2 = V1 + ΔV
V2 = 49.16 L + 0.2 L
V2 = 49.36 L
3. Calculate q (heat transferred):
Since the process is adiabatic (no heat transfer), q = 0.
4. Calculate w (work done):
The work done by the gas can be calculated using the formula:
w = PΔV
Here, ΔV is the change in volume and P is the external pressure (1 atm):
w = (1 atm)(0.2 L)
w = 0.2 L.atm
5. Calculate ΔU (change in internal energy):
ΔU = q - w
Since q = 0, ΔU = -w
ΔU = -0.2 L.atm
6. Calculate ΔT (change in temperature):
ΔT = ΔU / nCv,m
Here, n is the number of moles (2 moles) and Cv,m is the molar heat capacity at constant volume (28.8 J/mol.K).
ΔT = (-0.2 L.atm) / (2 moles)(28.8 J/mol.K)
ΔT = -0.00347 K
Since ΔT is very small, we can ignore it and assume the temperature remains constant.
7. Calculate ΔS (change in entropy):
ΔS = nCv,mln(T2 / T1)
Here, T1 is the initial temperature (25°C = 298 K) and T2 is the final temperature.
ΔS = (2 moles)(28.8 J/mol.K) ln(298 K / 298 K)
ΔS = 0
Therefore, ΔS = 0.
To summarize:
q = 0
w = 0.2 L.atm
ΔU = -0.2 L.atm
ΔT = -0.00347 K (approximately 0 K; negligible)
ΔS = 0
Note: The negative sign for ΔU indicates that the internal energy of the system has decreased.