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

The change in entropy of gas is
`\Delta S= nC_(P)ln3`

Step-by-step explanation:

n= Number of moles of gas

Change in entropy of gas =
`ds= \int (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)} (nC_(P)dT)/(T)`

`\Delta S = nC_(p)ln((T_(f))/(T_(i)))`

`\Delta S = nC_(p)ln3`

Therefore, The change in entropy of gas is
`\Delta S= nC_(P)ln3`