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In a cylinder–piston arrangement, 2kg of an ideal gas are expanded adiabatically from a temperature of 125°C to 30°C and it is found to perform 152KJ of work during the process while its enthalpy change is 212.8KJ. Find its specific heats at constant volume and constant pressure and characteristic gas constant. |
Answer» m = 2Kg T1 = 125°C T2 = 30°C W = 152KJ H = 212.8KJ CP = ?, CV = ?, R = ? We know that during adiabatic process is: W.D. = P1V1 – P2V2/γ–1 = mR(T1 – T2)/ γ–1 152 × 103 = 2 × R (125 – 30)/(1.4 – 1) R = 320J/Kg°K = 0.32 KJ/Kg°K H = mcp dT 212.8 = 2.CP.(125 – 30) CP = 1.12 KJ/Kg°K CP – CV = R CV = 0.8 KJ/Kg°K |
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