0.05 kg of carbon dioxide (molecular weight = 44) is compressed from 1 bar, 15°C, until
the pressure is 8.3 bar, and the volume is then 0.004 m3. Calculate the change of entropy.
Take cp for carbon dioxide as 1.0 kJ/kg K, and assume carbon dioxide to be a perfect gas (draw pc diagram )
0.05 kg of carbone dioxide (M=44) is compressed from I bar & 15C^° until the pressure is 8.3 bar and the volume is then 0.004m^3. Calculate the change of entropy
Answer
To calculate the change of entropy, we can use the equation:
ΔS = mcpln(T2/T1) - Rln(V2/V1)
Where:
ΔS = change of entropy
m = mass of carbon dioxide
cpl = specific heat capacity at constant pressure for carbon dioxide
T1 = initial temperature
T2 = final temperature
V1 = initial volume
V2 = final volume
R = specific gas constant for carbon dioxide
Given:
m = 0.05 kg
cpl = 1.0 kJ/kg K
T1 = 15°C = 273.15 + 15 K = 288.15 K
T2 = ?
V1 = ?
V2 = 0.004 m^3
R = R specific gas constant for carbon dioxide (value = 0.1889 kJ/kg K)
To find T2 and V1, we can use the ideal gas law:
PV = mRT
1 bar * V1 = 0.05 kg * 0.1889 kJ/kg K * 288.15 K
V1 = (0.05 kg * 0.1889 kJ/kg K * 288.15 K) / (1 bar)
V1 ≈ 2.73 m^3
Now, we can calculate T2:
8.3 bar * 0.004 m^3 = 0.05 kg * 0.1889 kJ/kg K * T2
T2 = (8.3 bar * 0.004 m^3) / (0.05 kg * 0.1889 kJ/kg K)
T2 ≈ 698.52 K
Now we have all the values to calculate the change of entropy:
ΔS = (0.05 kg * 1.0 kJ/kg K * ln(698.52 K / 288.15 K)) - (0.1889 kJ/kg K * ln(0.004 m^3 / 2.73 m^3))
Calculating this equation will give us the change of entropy.
To calculate the change in entropy (ΔS), we can use the equation:
ΔS = ∫(dq/T)
where dq is the heat transfer and T is the temperature. In this case, we are working with a compressible substance, so we need to consider the specific heat capacity of carbon dioxide (cp) and the compression process.
First, let's determine the initial and final temperatures since we need them to calculate the change in entropy. The temperature remains constant during the process, so both the initial and final temperatures are 15°C, which we need to convert to Kelvin (K):
Initial temperature (T1) = 15°C + 273.15 = 288.15 K
Final temperature (T2) = 15°C + 273.15 = 288.15 K
Next, we need to calculate the heat transfer (dq) during the process. Since we don't have any information about the heat transfer, we will assume it to be negligible. Therefore, dq = 0.
Now, we can proceed to calculate the change in entropy:
ΔS = ∫(dq/T)
Since dq is zero, the equation becomes:
ΔS = ∫(0/T) = 0
This means that the change in entropy is zero, which is expected for an isothermal compression process.
Regarding the pV diagram you mentioned, it is not possible to draw a specific pV diagram without knowing the specific details of the compression process (such as the number of stages, intermediate pressures, and volumes). However, I can explain the general characteristics of the pV diagram for an isothermal compression process.
In an isothermal compression process, the gas is compressed while maintaining the temperature constant. This means that the pressure and volume will have an inverse relationship, following Boyle's Law.
On a pV diagram, you would see a curve that starts at the initial pressure and volume and ends at the final pressure and volume. The curve would be a hyperbola, gradually curving upwards as the gas is compressed. The area below the curve represents the work done on the gas during the compression process.
I hope this explanation helps! Let me know if you have any further questions.