Chapter 20, Problem 15P

(a)

To determine

The final temperature.

Answer to Problem 15P

The final temperature is $380\text{ K}$ .

Explanation of Solution

The heat lost from the oxygen at higher temperature must be equal to the heat gained by oxygen at lower temperature.

Write the equation for the heat change of the system.

  $Q_{\text{cold}}=-Q{\text{hot}}_{}$                                                                                                         (I)

Here, $Q_{\text{cold}}$ is the heat gained by oxygen at lower temperature and $Q_{\text{hot}}$ is the heat lost from oxygen at higher temperature.

Write the equation for $Q_{\text{cold}}$ .

  $Q_{\text{cold}}=m_{\text{cold}}c\left(T-T_1\right)$                                                                                          (II)

Here, $m_{\text{cold}}$ is the mass of the oxygen at lower temperature, $c$ is the specific heat capacity of oxygen, $T$ is the final temperature and $T_1$ is the initial temperature of the oxygen at lower temperature.

The mass can be expressed as the product of molar mass and the number of moles of the gas.

Write the expression for $m_{\text{cold}}$ .

  $m_{\text{cold}}=n_{1}M$

Here, $n_1$ is the number of moles of oxygen at lower temperature and $M$ is the molar mass of oxygen.

Put the above equation in equation (II).

  $Q_{\text{cold}}=n_{1}Mc\left(T-T_1\right)$                                                                                         (III)

Write the equation for $Q_{\text{hot}}$ .

  $Q_{\text{hot}}=m_{\text{hot}}c\left(T-T_2\right)$                                                                                             (IV)

Here, $m_{\text{hot}}$ is the mass of oxygen at higher temperature and $T_2$ is the initial temperature of oxygen at higher temperature.

Write the expression for $m_{\text{hot}}$ .

  $m_{\text{hot}}=n_{2}M$

Here, $n_2$ is the number of moles of oxygen at higher temperature.

Put the above equation in equation (IV).

  $Q_{\text{hot}}=n_{2}Mc\left(T-T_2\right)$                                                                                             (V)

Put equations (III) and (V) in equation (I).

  $\begin{array}{c}{n}_{1}Mc\left(T-T_1\right)=-n_{2}Mc\left(T-T_2\right)\\ n_1\left(T-T_1\right)=-n_2\left(T-T_2\right)\end{array}$                                                                              (VI)

Write the ideal gas equation.

  $PV=nRT$

Here, $P$ is the pressure of the gas, $V$ is the volume of the gas, $R$ is the universal gas constant and $T$ is the temperature of the gas.

Rewrite the above equation for $n$ .

  $n=\frac{PV}{RT}$                                                                                                               (VII)

Use equation (VII) to write the expression for $n_1$ .

  $n_1=\frac{P_1{V}_1}{R{T}_1}$                                                                                                              (VIII)

Here, $P_1$ is the pressure of oxygen at lower temperature and $V_1$ is the volume of oxygen at lower temperature.

Use equation (VII) to write the expression for $n_2$ .

  $n_2=\frac{P_2{V}_2}{R{T}_2}$                                                                                                                (IX)

Here, $P_2$ is the pressure of oxygen at higher temperature and $V_2$ is the volume of oxygen at higher temperature.

Put equations (VIII) and (IX) in equation (VI) and rewrite it for $T$ .

  $\begin{array}{c}\frac{P_1{V}_1}{R{T}_1}\left(T-T_1\right)=-\frac{P_2{V}_2}{R{T}_2}\left(T-T_2\right)\\ \frac{P_1{V}_1}{T_1}T-\frac{P_1{V}_1}{T_1}{T}_1=-\frac{P_2{V}_2}{T_2}T+\frac{P_2{V}_2}{T_2}{T}_2\\ T\left(\frac{P_1{V}_1}{T_1}+\frac{P_2{V}_2}{T_2}\right)=P_1{V}_1+P_2{V}_2\\ T=\frac{P_1{V}_1+P_2{V}_2}{\left(\frac{P_1{V}_1}{T_1}+\frac{P_2{V}_2}{T_2}\right)}\end{array}$                                                                   (X)

Conclusion:

Substitute $1.75\text{ atm}$ for $P_1$ , $16.8\text{ L}$ for $V_1$ , $300\text{ K}$ for $T_1$ , $2.25\text{ atm}$ for $P_2$ , $22.4\text{ L}$ for $V_2$ and $450\text{ K}$ for $T_2$ in equation (X) to find $T$ .

  $\begin{array}{c}T=\frac{\left(1.75\text{ atm}\right)\left(16.8\text{ L}\right)+\left(2.25\text{ atm}\right)\left(22.4\text{ L}\right)}{\left(\frac{\left(1.75\text{ atm}\right)\left(16.8\text{ L}\right)}{300\text{ K}}+\frac{\left(2.25\text{ atm}\right)\left(22.4\text{ L}\right)}{450\text{ K}}\right)}\\ =380\text{ K}\end{array}$

Therefore, the final temperature is $380\text{ K}$ .

(b)

To determine

The final pressure.

Answer to Problem 15P

The final pressure is $2.04\text{ atm}$ .

Explanation of Solution

Rewrite the ideal gas equation for pressure.

  $P=\frac{nRT}{V}$

Use the above equation to write the expression for the final pressure of the oxygen.

  $P_f=\frac{n_{f}RT}{V_f}$                                                                                                             (XI)

Here, $n_f$ is the number of moles in the final state, $P_f$ is the final pressure and $V_f$ is the final volume.

For the final state of the system, the value of $n$ will be equal to the sum of $n_1$ and $n_2$ , the value of $V$ will be equal to the sum of $V_1$ and $V_2$ .

Write the equation for $n_f$ .

  $n_f=n_1+n_2$

Put equations (VIII) and (IX) in the above equation.

  $\begin{array}{c}{n}_f=\frac{P_1{V}_1}{R{T}_1}+\frac{P_2{V}_2}{R{T}_2}\\ =\frac{1}{R}\left(\frac{P_1{V}_1}{T_1}+\frac{P_2{V}_2}{T_2}\right)\end{array}$                                                                                            (XII)

Write the equation for $V_f$ .

  $V_f=V_1+V_2$                                                                                                        (XIII)

Put equations (X), (XII) and (XIII) in equation (XI).

  $\begin{array}{c}{P}_f=\left(\frac{1}{R}\left(\frac{P_1{V}_1}{T_1}+\frac{P_2{V}_2}{T_2}\right)\right)\frac{R}{V_1+V_2}\frac{P_1{V}_1+P_2{V}_2}{\left(\frac{P_1{V}_1}{T_1}+\frac{P_2{V}_2}{T_2}\right)}\\ =\frac{P_1{V}_1+P_2{V}_2}{V_1+V_2}\end{array}$                                                    (XIV)

Conclusion:

Substitute $1.75\text{ atm}$ for $P_1$ , $16.8\text{ L}$ for $V_1$ , $2.25\text{ atm}$ for $P_2$ and $22.4\text{ L}$ for $V_2$ in equation (XIV) to find $P_f$ .

  $\begin{array}{c}{P}_f=\frac{\left(1.75\text{ atm}\right)\left(16.8\text{ L}\right)+\left(2.25\text{ atm}\right)\left(22.4\text{ L}\right)}{16.8\text{ L}+22.4\text{ L}}\\ =2.04\text{ atm}\end{array}$

Therefore, the final pressure is $2.04\text{ atm}$ .