ThermodynamicsHard
Question
For the cyclic process given below, which of the following relations are correct?
Options
A.$\Delta S = S_{2} - S_{1} = \int_{1}^{2}\frac{\delta q_{rev}}{T}$
B.$\Delta S = S_{1} - S_{2} = \int_{2}^{1}\frac{\delta q_{irr}}{T}$
C.$\Delta S_{\text{cycle}} = 0 = \int_{1}^{2}{\frac{\delta q_{rev}}{T} +}\int_{2}^{1}\frac{\delta q_{irr}}{T}$
D.$\Delta S_{\text{cycle}} = 0 > \left( \int_{1}^{2}{\frac{\delta q_{rev}}{T} +}\int_{2}^{1}\frac{\delta q_{irr}}{T} \right)$
Solution
$dS = \frac{r_{rev}}{T}\text{ and }\oint_{}^{}{dS} = 0$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
The ΔG in the process of melting of ice at −15°C and 1 atm, is...Change in entropy is negative for...Internal energy does not include:...For a process to occur under adiabatic conditions, the essential condition(s) is/are...N2 + 3H2 ⇋ 2NH3 Which is correct statement if N2 is added at equilibrium condition?...