91Ó°ÊÓ

Two containers of a diatomic gas,

One container has a temperature increase of ${20}^{∘}C$

The other container receives the same quantity of heat energy, but at constant volume.

Knowing the formulas for the heat in isobaric and isochoric processes and that this heat is the same for both, just as the amount of gas, we can write

$n{C}_v(\mathrm{Δ}T{)}_v=Q=n{C}_p(\mathrm{Δ}T{)}_p$

From this equality we can simplify the number $n$of moles and express the temperature increase in the isochoric process as

$(\mathrm{Δ}T{)}_v=\frac{C_p}{C_v}(\mathrm{Δ}T{)}_p$

We should remember that the ratio of the isobaric heat capacity to the isochoric one is constant and is denoted by $\mathrm{γ}:=\frac{C_p}{C_v}$.

For diatomic gases, as is our case, we have $\mathrm{γ}=1.40$.

This means that we can substitute numerically to find,

$(\mathrm{Δ}T{)}_v=1.40·20=28\mathrm{K}$

Since we are talking about temperature difference, the unit will be measured in Celsius$28\mathrm{K}=28°\mathrm{C}.$