Question
Derive the relationship between $\text{C}_\text{p}$ and $\text{C}_\text{V}$ for an ideal gas.
✓
Answer
At constant volume $\text{C}_\text{V}=\Big(\frac{\text{dE}}{\text{dT}}\Big)_\text{V}$ for an ideal gas$\text{C}_\text{P}=\Big(\frac{\text{dH}}{\text{dT}}\Big)_\text{P}$ for an ideal gas
$\text{H}=\text{E}+\text{PV}=\text{E}+\text{RT}$
$\frac{\text{dH}}{\text{dt}}=\frac{\text{dE}}{\text{dt}}+\text{R}$
$\Rightarrow\text{C}_\text{P}=\text{C}_\text{V}+\text{R}$
$\Rightarrow\text{C}_\text{P}=\text{C}_\text{V}+\text{R}$ for an ideal gas
Alternate Answer
'q' at constant volume = $\text{q}_\text{v} $=\text{C}_\text{v}\Delta\text{T}=\Delta\text{U}$ 'q' at constant pressure = $\text{q}_\text{p} $=\text{C}_\text{p}\Delta\text{T}=\Delta\text{H}$$\Delta\text{H}=\Delta\text{U}+\Delta(\text{pV})$
$\Delta\text{H}=\Delta\text{U}+\Delta(\text{RT})$ $[\because\text{pV}=\text{RT}]$
$\Delta\text{H}=\Delta\text{U}+\text{R}\Delta\text{T}$ $[\because\text{‘R’} \text{ is constant}]$
$\text{C}_\text{p}\Delta\text{T}=\text{C}_\text{r}\Delta\text{T}+\text{R}\Delta\text{T}$
$\text{C}_\text{p}=\text{C}_\text{v}+\text{R}$
$\text{C}_\text{p}-\text{C}_\text{v}=\text{R}$
$\text{H}=\text{E}+\text{PV}=\text{E}+\text{RT}$
$\frac{\text{dH}}{\text{dt}}=\frac{\text{dE}}{\text{dt}}+\text{R}$
$\Rightarrow\text{C}_\text{P}=\text{C}_\text{V}+\text{R}$
$\Rightarrow\text{C}_\text{P}=\text{C}_\text{V}+\text{R}$ for an ideal gas
Alternate Answer
'q' at constant volume = $\text{q}_\text{v} $=\text{C}_\text{v}\Delta\text{T}=\Delta\text{U}$ 'q' at constant pressure = $\text{q}_\text{p} $=\text{C}_\text{p}\Delta\text{T}=\Delta\text{H}$$\Delta\text{H}=\Delta\text{U}+\Delta(\text{pV})$
$\Delta\text{H}=\Delta\text{U}+\Delta(\text{RT})$ $[\because\text{pV}=\text{RT}]$
$\Delta\text{H}=\Delta\text{U}+\text{R}\Delta\text{T}$ $[\because\text{‘R’} \text{ is constant}]$
$\text{C}_\text{p}\Delta\text{T}=\text{C}_\text{r}\Delta\text{T}+\text{R}\Delta\text{T}$
$\text{C}_\text{p}=\text{C}_\text{v}+\text{R}$
$\text{C}_\text{p}-\text{C}_\text{v}=\text{R}$
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