Chapter 6, Problem 6.60P

A ballonist begins a trip in a helium-filled balloon in early morning when the temperature is 15°C. By mid-afternoon, the temperature is 30.°C. Assuming the pressure remains at 1.00 atm, for each mole of helium, calculate the following:

1. The initial and final volumes
2. The change in internal energy, $\text{Δ}\text{E}$ (Hint: Helium behaves like an ideal gas, so $\text{E}=\frac{3}{2}n\text{R}\text{T}.$ Be sure the units of R are consistent with those of E.)
3. The work (w) done by the helium (in J)
4. The heat (q) transferred (in J) $\text{Δ}\text{H}$ for the process (in J)
5. Explain the relationship between the answers to parts (d) and (e).

Interpretation Introduction

Interpretation: The initial and final volume of the gas needs to be determined.

Concept Introduction : To determine the initial and final volume of gas, the ideal gas equation can be used.

  $P_1{V}_1=nR{T}_1$ and $P_2{V}_2=nR{T}_2$

Where, $P_1$ and $P_2$ are initial and final pressures, $V_1$ and $V_2$ are initial and final volumes and n is number of moles of gas, R is gas constant and $T_1$ and $T_2$ are initial and final temperatures respectively.

Answer to Problem 6.60P

The initial volume of gas is 23.6 L and the final volume of the gas is 24.9 L.

Explanation of Solution

Rearrange the ideal gas equation to solve $V_1$ and $V_2$ .

  $V_1=\frac{nR{T}_1}{P_1}$ and $V_2=\frac{nR{T}_2}{P_2}$

First calculate $V_1$ .

  $\begin{array}{l}{P}_1=1\text{atm}\\ n=1\text{mol}\\ R=0.0821\text{L}\text{atm}{\text{mol}}^{-1}{\text{K}}^{-1}\end{array}$

  $\begin{array}{c}{T}_1=15°\text{C}\\ \text{=15+273}\\ \text{=288}\text{K}\end{array}$

Put all the above values in the formula of $V_1$ .

  $\begin{array}{c}{V}_1=\frac{nR{T}_1}{P_1}\\ =\frac{1\text{mol}\times \text{0}\text{.0821}\text{L atm }{\text{mol}}^{-1}{\text{K}}^{-1}\times 288K}{1\text{atm}}\\ =23.6\text{L}\end{array}$

First calculate $V_2$ .

  $\begin{array}{l}{P}_2=1\text{atm}\\ n=1\text{mol}\\ R=0.0821\text{L}\text{atm}{\text{mol}}^{-1}{\text{K}}^{-1}\end{array}$

  $\begin{array}{c}{T}_2=30°\text{C}\\ \text{=30+273}\\ \text{=303}\text{K}\end{array}$

Put all the above values in the formula of $V_2$ .

  $\begin{array}{c}{V}_2=\frac{nR{T}_2}{P_2}\\ =\frac{1\text{mol}\times \text{0}\text{.0821}\text{L atm }{\text{mol}}^{-1}{\text{K}}^{-1}\times 303K}{1\text{atm}}\\ =24.9\text{L}\end{array}$

The final volume of the gas will be 24.9 L.

Interpretation Introduction

Interpretation: The internal energy needs to be calculated.

Concept Introduction : The formula to calculate change of internal energy is as follows:

  $\Delta E=\frac{3}{2}nR\Delta T$

Here, $\Delta E$ is change in internal energy, n is number of moles of gas, R is gas constant and $\Delta T$ is change of temperature.

Answer to Problem 6.60P

Value of $\Delta E$ is $\text{187}\text{J}$ .

Explanation of Solution

First determine the change in temperature $\left(\Delta T\right)$

as follows:

  $\begin{array}{c}\Delta T=T_2-T_1\\ =303\text{K}-288\text{K}\\ \text{=15}\text{K}\end{array}$

Now use the formula to calculate $\Delta E$ .

  $\begin{array}{c}\Delta E=\frac{3}{2}nR\Delta T\\ =\frac{3}{2}\times 1\text{mol}\times 8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times 15\text{K}\\ \text{=187}\text{J}\end{array}$

Interpretation Introduction

Interpretation: The work done ( w ) by helium needs to be determined.

Concept Introduction : Work done is negative when it is done by the system. Determine the work done using the following equation,

  $w=-P\Delta V$

Where, w is work done, P is pressure and $\Delta V$ is change in volume.

Answer to Problem 6.60P

Work done is $-1.3\times {10}^2\text{J}$ .

Explanation of Solution

Given Information:

  $\begin{array}{l}P=1\text{atm}\\ {\text{V}}_2=24.9\text{L}\\ {\text{V}}_1=23.6\text{L}\end{array}$

  $\begin{array}{c}\Delta V=24.9\text{L}-23.6\text{L}\\ \text{=1}\text{.3}\text{L}\end{array}$

Calculation:

Put all the values in the below equation.

  $\begin{array}{c}w=-P\Delta V\\ =-1\text{atm}\times 1.3\text{L}\\ \text{=}-\text{1}\text{.3}\text{atm}\text{L}\times \frac{1\text{J}}{9.87\times {10}^{-3}\text{atm L}}\left(1\text{J=9}\text{.87}\times {\text{10}}^{-3}\text{atm}\right)\\ =-1.3\times {10}^2\text{J}\end{array}$

Hence, work done is $-1.3\times {10}^2\text{J}$ .

Interpretation Introduction

Interpretation: The heat transferred in joule needs to be determined.

Concept Introduction: The relation between internal energy $\left(\Delta E\right)$ , heat ( q ) and work done( w ) is,

  $\Delta E=q+w$

Answer to Problem 6.60P

Heat transferred in joule is 317 J.

Explanation of Solution

  $\begin{array}{l}w=-1.3\times {10}^2\text{J}\\ \Delta E=187\text{J}\end{array}$

Put the above values in the equation below to calculate q .

  $\Delta E=q+w$

Rearrange the above equation to calculate q .

  $\begin{array}{c}q=\Delta E-w\\ =187\text{J}-(-1.3\times {10}^2\text{J})\\ =317\text{J}\end{array}$

Interpretation Introduction

Interpretation: The value of $\Delta H$ for the process in joule needs to be determined.

Concept Introduction: At constant pressure, enthalpy change $\left(\Delta H\right)$ is equal to $\left(q_{\text{p}}\right)$ .

  $\Delta H=q_{\text{p}}$

Answer to Problem 6.60P

Enthalpy change of the system is 317 J.

Explanation of Solution

At constant pressure, enthalpy change $\left(\Delta H\right)$ is equal to $\left(q_{\text{p}}\right)$ .

  $\begin{array}{c}\Delta H=q_{\text{p}}\\ =317\text{J}\end{array}$

Interpretation Introduction

Interpretation: The relationship between the answers to parts (d) and (e) needs to be explained.

Concept Introduction: From part (d), The relation between internal energy $\left(\Delta E\right)$ , heat ( q ) and work done( w ) is,

  $\Delta E=q+w$

From part (e), At constant pressure, enthalpy change $\left(\Delta H\right)$ is equal to $\left(q_{\text{p}}\right)$ .

  $\Delta H=q_{\text{p}}$

Answer to Problem 6.60P

It is proved that at constant pressure, change of enthalpy is equal to heat .

Explanation of Solution

Enthalpy change $\Delta H$ and internal energy change $\Delta E$ is related as follows:

  $\begin{array}{c}\Delta H=\Delta E+P\Delta V\\ =q+w+P\Delta V(\Delta E=q+w)\\ =q-P\Delta V+P\Delta V(w=-P\Delta V)\\ =q_p(\text{At}\text{constant}\text{pressure})\end{array}$