An engine containing 2 kg air as working substance is initially at 1 atm and 27 0C. The system undergoes isochoric process to a point where pressure of the system is 2 atm. At this point heat is transferred to the air untill the volume doubles. Calulate the total workdone and the amount of heat transferred to the air.
Answers
Explanation:
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Concept:
The isochoric process is the process in which the volume is constant. As for the isochoric process, volume is constant, and the work done is zero.
Given:
Mass, m = 2 kg
Pressure, P₁ = 1 atm
Temperature, T₁ = 27°C = (27+273)K = 300 K
Pressure, P₂ = 2 atm
The volume doubled as heat is applied.
Find:
The total Work done, W and the Heat transferred, Q.
Solution:
According to the isochoric process, when the volume is constant, the pressure applied to the gas is directly proportional to the temperature of the gas.
Final temperature,
T₂ = T₁ × P₂/P₁
T₂ = 300 × (2/1) = 600K
Total work done to the air is zero because the process is isochoric.
The heat transferred to the air can be calculated by the temperature difference of air.
Heat transferred to the air, Q = mCv(T₂- T₁ )
Specific heat of air, Cv= 0.718 kJ/kg.K
Q = mCv(T₂- T₁ ) =2 × 0.718(600 - 300) = 430.800 kJ
Hence, the total work done is zero, and the heat transferred to the air is 430.8 kJ.
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