Question
During an adiabatic process, the temperature of 6.50 moles of a monatomic ideal gas drops from 525 °C to 169 °C. What is the work done?
Here is my work, is it correct?
T1 = 525+273 = 798 K
T2 = 169 + 273 = 442 K
work done = R[T2-T1]/1-gamma = 8.314[525 - 798]/(1-1.67) = - 4417.58 J
Here is my work, is it correct?
T1 = 525+273 = 798 K
T2 = 169 + 273 = 442 K
work done = R[T2-T1]/1-gamma = 8.314[525 - 798]/(1-1.67) = - 4417.58 J