Question

When$\mathbf{1}\mathbf{.}\mathbf{0}\mathit{m}\mathit{o}\mathbf{\text{l}}$of oxygen (${\mathit{O}}_{\mathbf{2}}$) gas is heated at constant pressure starting at,$\mathbf{0}\mathbf{°}\mathit{C}$ how much energy must be added to the gas as heat to double its volume? (The molecules rotate but do not oscillate.)

Short answer

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Step-by-step solution

Given data

Number of moles of gas is$n=1.0$

Temperature;

$\begin{array}{c}{\text{T}}_i=0°\text{C}\\ =273\text{\hspace{0.17em}K}\end{array}$

Understanding the concept

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, $C_P$is the specific heat capacity at constant pressure and $\Delta T$is the temperature difference.

Calculate the amount of energy that must be added to the gas as heat to double its volume

The equation;

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

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

Substitute $\frac{7}{2}R$for$C_P$

$Q=\frac{7}{2}nR\Delta T$....... (i)

At constant pressure, Boyle’s law gives,

$$\begin{gathered} \text{T}\propto \text{V} \\ \therefore \frac{T_f}{T_i}=\frac{V_f}{V_i} \\ {\text{T}}_f=2{\text{T}}_i \\ \begin{array}{c}\Delta T=T_f-T_i\\ =2T_i-T_i\\ =T_i\end{array} \end{gathered}$$

Substitute$1mol$for n, $8.3145J/mol\cdot K$ for R and $273K$for$\Delta T$ into the equation (i)

$\begin{array}{c}Q=\frac{7}{2}\left(1\right)\left(8.3145\right)\left(273\right)\\ =7.945\times {10}^3\text{J}\\ \approx 8.0\times {10}^3\text{J}\end{array}$

Therefore, the amount of energy that must be added to the gas as heat to double its volume is$8.0\times {10}^3\text{J}.$