Question

A sample consisting of n mol of an ideal gas undergoes a reversible isobaric expansion from volume Vi to volume 3Vi. Find the change in entropy of the gas by calculating, ∫dQ / T, where dQ = nCPdT. (Use the following as necessary: Cp and n.)

Answer 1

Answer:

The change in entropy of gas is $\Delta S= nC_{P}ln3$

Explanation:

n= Number of moles of gas

Change in entropy of gas = $ds= \int \frac{dQ}{T}$

$dQ= nC_{p}dT$

From the given,

$V_{i}=V$

$V_{f}=3V$

Let "T" be the initial temperature.

$\frac {V_{i}}{T_{i}}=\frac {V_{f}}{T_{f}}$

$\frac {V}{T}=\frac {3V}{T_{f}}$

${T_{f}} = 3T$

$\int ds = \int ^{T_{f}}_{T_{i}} \frac{nC_{P}dT}{T}$

$\Delta S = nC_{p}ln(\frac{T_{f}}{T_{i}})$

$\Delta S = nC_{p}ln3$

Therefore, The change in entropy of gas is $\Delta S= nC_{P}ln3$