Question

An experimenter adds of 970J heat to 1.75mol of an ideal gas to heat it from$\mathbf{10}\mathbf{°}\mathit{C}$ to$\mathbf{25}\mathbf{°}\mathit{C}$ at constant pressure. The gas does + 223J of work during the expansion. (a) Calculate the change in internal energy of the gas. (b) Calculate$\mathit{\gamma }$ for the gas.

Short answer

1. $\Delta U=747J$
   
2. $\gamma =1.3Q$
   

Step-by-step solution

Equations involved

The first law of thermodynamics is $\mathit{\Delta }\mathit{U}\mathbf{=}\mathit{Q}\mathbf{-}\mathit{W}$.

The ratio of heat capacities is given by$\mathit{\gamma }\mathbf{=}\frac{{\mathbf{C}}_{\mathbf{P}}}{{\mathbf{C}}_{\mathbf{V}}}$.

The heat energy is calculated by$\mathit{Q}\mathbf{=}\mathit{n}{\mathit{C}}_{\mathbf{P}}\mathit{\Delta }\mathit{T}$.

The relation between the heat capacities is given by${\mathit{C}}_{\mathbf{P}}\mathbf{-}{\mathit{C}}_{\mathbf{V}}\mathbf{=}\mathit{R}$.

Calculate the internal energy

Given that

$$\begin{gathered} Q=970JandW=223J. \\ \Delta U=Q-W \\ \Delta U=970-223 \\ \Delta U=747J \end{gathered}$$

Calculate the heat capacities

We know that$Q=n{C}_P\Delta T$.

$$\begin{gathered} C_P=\frac{Q}{n\Delta T} \\ {C}_P=\frac{970}{1.75\times 15}/ \\ {C}_P=37J/mol.K \end{gathered}$$

Now,$C_P-C_V=R$
$$\begin{gathered} C_V=C_P-R \\ {C}_V=37-8.314 \\ {C}_V=28.68J/mol.K \end{gathered}$$

Calculate the ratio of heat capacities

$$\begin{gathered} \gamma =\frac{C_P}{C_V} \\ \gamma =\frac{37}{28.68} \\ \gamma =1.3 \end{gathered}$$