91Ó°ÊÓ

Mass of oxygen is m=12.0g

Temperature;

$$\begin{gathered} T_i=25.0^{∘}C \\ {T}_f=125.0^{∘}C \end{gathered}$$

The expression for the amount of energy transferred to the gas is given by,

$Q=n{C}_P\mathrm{Δ}T$

Here Q is the amount of energy transferred to gas, n is the number of moles, C_Pis the specific heat capacity at constant pressure and$\mathrm{Δ}T$is the temperature difference.

The expression for the change in internal energy is given by,

$\mathrm{Δ}{E}_{int}=n{C}_V\mathrm{Δ}T$

Here $\mathrm{Δ}{E}_{int}$is the change in internal energy, n is the number of moles, C_vis the specific heat capacity at constant volume,$\mathrm{Δ}T$ is the difference in temperature.

We know that,

$Numberofmoles\left(n\right)=\frac{Massofthesample}{Molarmassofthesample}$

It is given that the mass of the sample is 12.0g .

Now the molar mass of the oxygen (O₂ )molecule is 32g .

$\begin{array}{l}n=\frac{12.0}{32.0}\\ =0.375mol\end{array}$

Therefore, the number of moles of oxygen is 0.375 mol.

Using the equation of amount of energy transferred to the gas,

$Q=n{C}_P\mathrm{Δ}T$…… (i)

But, for diatomic gas,$C_P==\frac{7}{2}R$

Substitute $\frac{7}{2}R$for C_P into the equation (i)

localid="1662466707322" $Q=\frac{7}{2}nR\mathrm{Δ}T$

Substitute 0.375mol for n,8.3145J/mol.K for R,$(125.0^{∘}C-{25}^{∘}C)$for$\mathrm{Δ}T$into the above equation, localid="1662614315121" $\begin{array}{l}Q=\frac{7}{2}(0.375)\left(8.314\left)\right(125-25\right)\\ =1091\\ =109×{10}^3]\end{array}$

Therefore, the amount of energy transferred to the oxygen as heat is 109$×$10³J.

The equation for the change in internal energy is given by,

$\mathrm{Δ}{E}_{int}=n{C}_V\mathrm{Δ}T$…… (ii)

But, for diatomic gasrole="math" localid="1662467172844" $C_V=\frac{5}{2}R$

Substitute 5/2Rfor C_Vinto the equation (ii)

$\mathrm{Δ}{E}_{int}=\frac{5}{2}nR\mathrm{Δ}T$

Substitute$0.375mol$for n,$8.3145J/molâ‹\dots K$for R, $(125°C-25°C)$for$\mathrm{Δ}T$into the above equation,

$$\begin{gathered} \mathrm{Δ}{E}_{int}=\frac{5}{2}\left(0.375\right)\left(8.314\right)\left(125-25\right) \\ =779.44J \end{gathered}$$

The ratio of the internal energy to the energy transferred to the oxygen

$\frac{\mathrm{Δ}{E}_{int}}{Q}=\frac{779.44}{1091}$

=0.714

Therefore, the fraction of the heat that is used to raise the internal energy of the oxygen is 0.714.