consider a system consisting of 2.0mol co2(g) initially at 25c and 10atm and confined to a cylinder of cross section 10.0cm2. it is allowed to expand adiabatically against an external pressure of 1.0atm until the piston has moved outwards through a 20cm. assume that co2 is a perfect gas with C vm=28.8J/K/MOL and calculate q,w,u,t,s
3.6(a) Consider a system consisting of 2.0 mol CO2(g), initially at 25°C and
10 atm and confined to a cylinder of cross-section 10.0 cm2
. It is allowed to
expand adiabatically against an external pressure of 1.0 atm until the piston
has moved outwards through 20 cm. Assume that carbon dioxide may be
considered a perfect gas with CV,m = 28.8 J K−1
mol
−1
and calculate (a) q, (b) w,
(c) ΔU, (d) ΔT, (e) ΔS.
Well, well, well, looks like we have quite the party going on here in this system! Time to crunch some numbers and get this fiesta started.
First, let's take a look at what we know: we have 2.0 moles of CO2, initial temperature of 25°C (which is 298 K), initial pressure of 10 atm, and a cylinder with a cross-section of 10.0 cm^2. Oh, and we have an adiabatic expansion against an external pressure of 1.0 atm, with the piston moving outwards by 20 cm. Hold on tight, we're diving into some calculations now!
To calculate q (heat transfer), we use the equation q = 0 for an adiabatic process since no heat is transferred.
Next up, we calculate w (work done). The work done during an adiabatic expansion can be calculated using the formula w = nCvm(T2 - T1), where n is the number of moles, Cvm is the molar heat capacity at constant volume, and T2 and T1 are the final and initial temperatures, respectively. Plugging in our numbers, we get w = 2.0 mol * 28.8 J/(K*mol) * (T2 - 298 K). But wait, we need T2! Let's use the ideal gas law to find it.
Using the ideal gas law equation PV = nRT, we can solve for T2. Here we go: (10 atm * 10.0 cm^2) * (20 cm) = 2.0 mol * R * T2. Solving for T2 gives us T2 = (2000 cm^3 * atm) / (2.0 mol * R), where R is the ideal gas constant. Plugging this value of T2 back into the w equation will give us the answer for w.
Now, u (internal energy). In an adiabatic expansion, the internal energy does not change because no heat is transferred. So, u = 0.
Finally, t (temperature change) and s (entropy change). In an adiabatic process, we have the relationship (T2/T1)^(Cvm/R) = (V1/V2)^(Cvm/R), where V1 and V2 are the initial and final volumes, respectively. We know the initial volume (10.0 cm^2 * 20 cm), and we can find the final volume by using the equation V2 = V1 * (P1/P2)^(1/gamma), where gamma is the heat capacity ratio (Cp/Cv) for CO2. Plugging in these values into the equation will give us the temperature change, t = T2 - T1, and the entropy change, s = Cvm * ln(T2/T1). Yay, we're almost done!
Alright, now it's time to put on my math hat and calculate all these things. *drumroll* Remember, folks, this is gonna be a wild ride with numbers flying around. Brace yourselves!
To calculate the values for q (heat), w (work), u (internal energy), t (temperature change), and s (entropy change), we can use the first law of thermodynamics and the ideal gas law.
1. Heat (q):
Since the expansion is adiabatic, there is no heat exchanged with the surroundings (q=0).
2. Work (w):
The work done during the expansion can be calculated using the equation:
w = -P_external * ΔV
where P_external is the external pressure and ΔV is the change in volume.
Given:
P_external = 1.0 atm = 1.01325 x 10^5 Pa
ΔV = A * Δx
A = cross-sectional area = 10.0 cm^2 = 10.0 x 10^-4 m^2
Δx = 20.0 cm = 20.0 x 10^-2 m
Using the given values, we can calculate:
ΔV = (10.0 x 10^-4 m^2) * (20.0 x 10^-2 m)
= 2.0 x 10^-4 m^3
w = -(1.01325 x 10^5 Pa) * (2.0 x 10^-4 m^3)
(Note: The negative sign indicates work done by the system.)
3. Internal Energy (u):
The change in internal energy (ΔU) can be calculated using the equation:
ΔU = q + w
Since q = 0 (adiabatic process), ΔU = w.
4. Temperature Change (t):
The temperature change can be calculated using the equation:
t = ΔU / (C_vm * n)
where C_vm is the molar heat capacity and n is the number of moles.
Given:
C_vm = 28.8 J/K/mol
n = 2.0 mol
ΔU = w
5. Entropy Change (s):
The entropy change can be calculated using the equation:
ΔS = C_vm * ln(T2 / T1) - R * ln(V2 / V1)
where:
T1 and T2 are the initial and final temperatures respectively,
V1 and V2 are the initial and final volumes respectively,
and R is the gas constant.
Given:
T1 = 25 °C = 25 + 273.15 = 298.15 K
T2 = ?
V1 = ?
V2 = (10.0 cm^2) * (20.0 cm) = 200 cm^3 = 200 x 10^-6 m^3
R = 8.314 J/(mol K)
Now, we have all the necessary information to calculate q, w, u, t, and s.
To find the values of q, w, u, T, and ΔS, we need to consider the changes in the system during the adiabatic expansion of CO2 gas. Let's break down each parameter step by step:
1. Calculate the work done (w):
The work done can be calculated using the formula: w = P_ext * ΔV. Since the process is adiabatic, there is no heat transfer (q = 0). The external pressure (P_ext) is given as 1.0 atm, and the change in volume (ΔV) can be calculated using the formula: ΔV = A * Δx, where A is the cross-sectional area and Δx is the change in piston position. In this case, A = 10.0 cm^2 and Δx = 20 cm. Convert the area and change in position to appropriate units and calculate ΔV.
2. Calculate the change in internal energy (ΔU):
The change in internal energy is given by the formula: ΔU = q - w. Since q = 0, ΔU can be equated to -w.
3. Calculate the final temperature (T_final):
The final temperature can be calculated using the adiabatic expansion formula: T_final = T_initial * (P_final / P_initial)^((γ-1)/γ), where T_initial is the initial temperature, P_final is the final pressure, P_initial is the initial pressure, and γ is the heat capacity ratio (C_p / C_V). In this case, T_initial = 25°C (converted to Kelvin), and P_initial = 10 atm, while P_final is given as 1 atm. Use the given value of γ = C_p / C_V = 28.8 J/K/mol to calculate T_final.
4. Calculate the change in entropy (ΔS):
The change in entropy can be calculated using the formula: ΔS = n * C_v * ln(T_final / T_initial), where n is the number of moles, C_v is the molar heat capacity at constant volume, and ln represents the natural logarithm. In this case, n = 2.0 mol, C_v = 28.8 J/K/mol, and T_final and T_initial are given.
5. Calculate the heat transfer (q):
Since the process is adiabatic, there is no heat transfer (q = 0).
After calculating these values, substitute the appropriate parameters and perform the calculations to find the values of q, w, ΔU, T_final, and ΔS.