Question

Match the following:
Thermodynamic Process $\left(\Delta S_{system}\right)$

Answer 1

We know that $\Delta S=\int \frac{d{q}_{rev.}}{T}$

Option A: Since isothermal, dU = 0, dq = -dW.

$\Delta S=\int \frac{PdV}{T}=\int \frac{nRdV}{V}=nRln\left(\frac{V_2}{V_1}\right)$

Option B: Since adiabatic irreversible, consider it a combination of reversible isothermal and reversible isochoric, hence

$\Delta S=nRln\left(\frac{V_2}{V_1}\right)+n{c}_{v}ln\left(\frac{T_2}{T_1}\right)$

Option C: Since isochoric, dW = 0. dQ = dU

$\Delta S=\int \frac{n{c}_{v}dT}{T}=n{c}_{v}ln\left(\frac{T_2}{T_1}\right)$

Option D: Since isobaric, dQ = dH

$\Delta S=\int \frac{n{c}_{p}dT}{T}=n{c}_{p}ln\left(\frac{T_2}{T_1}\right)$