Question

A balloon filled with 39.1 moles of helium has a volume of 876 L at$\mathbf{0}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}\mathbf{C}$and 1.00 atm pressure. At constant pressure, the temperature of the balloon is increased to $\mathbf{38}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}\mathbf{C}$,causing the balloon to expand to a volume of 998 L. Calculate q, w, and ∆E for the helium in the balloon. (The molar heat capacity for Helium gas is $\mathbf{20}\mathbf{.}\mathbf{8}\mathbf{\text{J}}\mathbf{°}{\mathbf{\text{C}}}^{\mathbf{-}\mathbf{1}}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}$.)

Short answer

Heat Energy (q) = 30.9 K J

Work done (w) = -12.4 K J

Change in Internal Energy = 18.5 J

Step-by-step solution

Calculation of Heat Energy (q):

Given,

The molar heat capacity,

$$\begin{gathered} \left({\mathrm{C}}_{\mathrm{m}}\right)=20.8\mathrm{J}/°\mathrm{C}\mathrm{mol} \\ =0.0208\mathrm{kJ}/°\mathrm{C}\mathrm{mo}/ \end{gathered}$$

Change in temperature,

$$\begin{gathered} \mathrm{\Delta T}={\mathrm{T}}_2-{\mathrm{T}}_1 \\ \mathrm{\Delta T}=38.08-0.08 \\ =38.00°\mathrm{C} \end{gathered}$$

The molar heat capacity,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{m}}=\frac{\mathrm{\Delta q}}{\mathrm{n\Delta T}} \\ =0.0208\mathrm{kJ} \\ 0.0208=\frac{△\mathrm{q}}{39.1\times 38.0} \\ △\mathrm{q}=30.9\mathrm{kJ} \\ \left(\mathrm{cm}\right)=\raisebox{1ex}{$\mathrm{nq}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{n}∆\mathrm{t}$}\right. \\ 0.0208\mathrm{K}\mathrm{J}==\frac{\mathrm{\Delta q}}{39.1\times 38.0} \\ \mathrm{q}=30.9\mathrm{K}\mathrm{J} \end{gathered}$$Hence, the heat energy is 30.9 KJ.

Calculation of Work done (w)

$$\begin{gathered} \mathrm{Workdone}\left(\mathrm{w}\right)=-\mathrm{P\Delta V} \\ =-1\mathrm{atm}\times 122\mathrm{L} \\ =-122\mathrm{Latm} \\ =12.4\mathrm{kJ} \end{gathered}$$

Calculation of Energy Change

$$\begin{gathered} \mathrm{\Delta E}=\mathrm{q}+\mathrm{w} \\ =30.9+(-12.4) \\ =18.5\mathrm{kJ} \end{gathered}$$

Thus, the energy change is found as 18.5 kJ.