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Answer :
A 2 kg mass of oxygen expands at a constant pressure of 172 kPa in piston-cylinder system from a temperature of 32° C to a final temperature of 182°C.
(a) The heat required is 552 kJ.
(b) The work done is 136.704 kJ.
(c) The change in enthalpy is 415.296 kJ.
(d) The change in internal energy is 278.592 kJ.
To determine the heat required, work, change in enthalpy, and change in internal energy for the given expansion of oxygen, we can use the first law of thermodynamics and the ideal gas law.
- Mass of oxygen (m): 2 kg
- Initial temperature (T1): 32°C
- Final temperature (T2): 182°C
- Constant pressure (P): 172 kPa
Step 1: Convert temperatures from Celsius to Kelvin.
T1_K = T1 + 273.15
T2_K = T2 + 273.15
Step 2: Calculate the heat required using the formula:
Q = m * C_p * (T2 - T1)
where m is the mass and C_p is the specific heat capacity at constant pressure.
The specific heat capacity of oxygen at constant pressure is approximately 0.92 kJ/(kg·K).
Q = 2 kg * 0.92 kJ/(kg·K) * (182 - 32)
Q = 2 kg * 0.92 kJ/(kg·K) * 150
Q = 552 kJ
Therefore, the heat required for the expansion is 552 kJ.
Step 3: Calculate the work done using the formula:
W = P * ΔV
where P is the constant pressure and ΔV is the change in volume.
To calculate ΔV, we can use the ideal gas law:
P * V = m * R * T
where R is the specific gas constant for oxygen (0.287 kJ/(kg·K)).
Rearranging the equation for volume:
V = (m * R * T) / P
ΔV = V2 - V1
= [(2 kg * 0.287 kJ/(kg·K) * T2_K) / P] - [(2 kg * 0.287 kJ/(kg·K) * T1_K) / P]
= (2 * 0.287 * (T2_K - T1_K)) / P
ΔV = (2 * 0.287 * (182 + 273.15 - 32 - 273.15)) / 172
ΔV = 0.792 m³
W = P * ΔV
W = 172 kPa * 0.792 m³
W = 136.704 kJ
Therefore, the work done during the expansion is 136.704 kJ.
Step 4: Calculate the change in enthalpy (ΔH) using the formula:
ΔH = Q - P * ΔV
ΔH = 552 kJ - 172 kPa * 0.792 m³
ΔH = 552 kJ - 136.704 kJ
ΔH = 415.296 kJ
Therefore, the change in enthalpy is 415.296 kJ.
Step 5: Calculate the change in internal energy (ΔU) using the equation:
ΔU = ΔH - P * ΔV
ΔU = 415.296 kJ - 172 kPa * 0.792 m³
ΔU = 415.296 kJ - 136.704 kJ
ΔU = 278.592 kJ
Therefore, the change in internal energy is 278.592 kJ.
To summarize:
(a) The heat required is 552 kJ.
(b) The work done is 136.704 kJ.
(c) The change in enthalpy is 415.296 kJ.
(d) The change in internal energy is 278.592 kJ.
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