Question

consider two quasistatic processes that take a sample of an ideal gas containing n particles from an initial state (p_i, v_i) to a final state (p_f, v_f). process a is a one - step isentropic compression. process b has two steps: an isochoric pressure increase to an intermediate state 1 followed by an isobaric decrease in volume. express the changes in entropy for process a and process b in terms of the given quantities.
b⇒ isochoric + isobaric
c⇒ isobaric + isochoric
δs(a)=0 : cyclic
δs(b)>0
δs(c)>0
adiabatic φ = 0
δs=∫_i^f\frac{dq}{t}
δs→0

Explanation:

Step1: Recall entropy - change formula for ideal gas

For an ideal gas, the entropy change $\Delta S$ is given by $\Delta S=\int\frac{dQ}{T}$. For an isentropic process, $dQ = 0$, so $\Delta S=0$. Process A is a one - step isentropic compression, so $\Delta S_A = 0$.

Step2: Analyze process B

Process B has two steps: an isochoric (constant - volume) pressure increase and an isobaric (constant - pressure) volume decrease. For an isochoric process, $dQ = nC_VdT$ and for an isobaric process $dQ=nC_pdT$. Since heat is exchanged in both steps and the process is not cyclic in a way that would cancel out the entropy changes, $\Delta S_B>0$.

Step3: Analyze process C

Although not fully described in the problem statement, if we assume it also involves heat exchange as it is not isentropic, and from the notes $\Delta S_C>0$.

Answer:

$\Delta S_A = 0$, $\Delta S_B>0$, $\Delta S_C>0$