Suppose 12.0gof oxygen (O₂) gas is heated at constant atmospheric pressure from$\mathbf{25}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}\mathbf{C}\mathbf{}\mathbf{to}\mathbf{}\mathbf{125}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}\mathbf{C}$to.

1. How many moles of oxygen are present?
2. How much energy is transferred to the oxygen as heat? (The molecules rotate but do not oscillate).
3. What fraction of heat is used to raise the internal energy of the oxygen?

Mass of oxygen is m=12.0g

Temperature;

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

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

$Q=n{C}_P\Delta T$

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$\Delta T$is the temperature difference.

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

$\Delta E_{int}=n{C}_V\Delta T$

Here $\Delta E_{int}$is the change in internal energy, n is the number of moles, C_vis the specific heat capacity at constant volume,$\Delta T$ 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,

$Q=n{C}_P\Delta T$…… (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" $Q=\frac{7}{2}nR\Delta T$

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

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

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

$\Delta E_{int}=n{C}_V\Delta T$…… (ii)

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

Substitute 5/2Rfor C_Vinto the equation (ii)

$\Delta E_{int}=\frac{5}{2}nR\Delta T$

Substitute$0.375mol$for n,$8.3145J/mol\cdot K$for R, $(125°C-25°C)$for$\Delta T$into the above equation,

$$\begin{gathered} \Delta E_{int}=\frac{5}{2}\left(0.375\right)\left(8.314\right)\left(125-25\right) \\ =779.44J \end{gathered}$$

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

$\frac{\Delta E_{int}}{Q}=\frac{779.44}{1091}$

=0.714

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