Chapter 15, Problem 18P

(a)

To determine

The heat absorbed by the gas.

Answer to Problem 18P

The heat absorbed by the gas is $\underset{\_}{436\text{J}}$.

Explanation of Solution

Given that the number of moles of the oxygen gas is $1.00\text{mol}$, the pressure is $1.00\text{atm}$, the initial temperature is $10.0\text{°C}$ ($283.15\text{K}$), and the final temperature is $25.0\text{°C}$ ($298.15\text{K}$).

Write the expression for the heat absorbed by the gas.

$Q=n{C}_{\text{p}}\Delta T$ (I)

Here, $Q$ is the heat, $n$ is the number of moles, $C_{\text{p}}$ is the molar specific heat at constant pressure, and $\Delta T$ is the change in temperature.

Write the expression for the molar specific heat capacity according to Mayer’s law.

$C_{\text{p}}=C_{\text{v}}+R$ (II)

Here, $C_{\text{v}}$ is the molar specific heat at constant volume, and $R$ is the universal gas constant.

Use equation (II) in (I) and expand the term $\Delta T$

$Q=n\left(C_{\text{v}}+R\right)\left(T_{\text{f}}-T_{\text{i}}\right)$ (III)

Here, $T_{\text{f}}$ is the final temperature, and $T_{\text{i}}$ is the initial temperature.

Since oxygen is a diatomic gas, the molar specific heat at constant volume is given by,

$C_{\text{v}}=\frac{5}{2}R$ (IV)

Use equation (IV) in (III).

$\begin{array}{c}Q=n\left(\frac{5}{2}R+R\right)\left(T_{\text{f}}-T_{\text{i}}\right)\\ =\frac{7}{2}nR\left(T_{\text{f}}-T_{\text{i}}\right)\end{array}$ (V)

Conclusion:

Substitute $1.00\text{mol}$ for $n$, $8.314\text{J/}\left(\text{mol}\cdot \text{K}\right)$ for $R$, $298.15\text{K}$ for $T_{\text{f}}$, and $283.15\text{K}$ for $T_{\text{i}}$ in equation (V) to find $Q$.

$\begin{array}{c}Q=\frac{7}{2}\left(1.00\text{mol}\right)\left[8.314\text{J/}\left(\text{mol}\cdot \text{K}\right)\right]\left(298.15\text{K}-283.15\text{K}\right)\\ =436.4\text{J}\\ \approx 436\text{J}\end{array}$

Therefore, the heat absorbed by the gas is $\underset{\_}{436\text{J}}$.

(b)

To determine

The change in volume of the gas in the process.

Answer to Problem 18P

The change in volume of the gas in the process is $\underset{\_}{1.23\text{L}}$.

Explanation of Solution

Given that the number of moles of the oxygen gas is $1.00\text{mol}$, the pressure is $1.00\text{atm}$, the initial temperature is $10.0\text{°C}$ ($283.15\text{K}$), and the final temperature is $25.0\text{°C}$ ($298.15\text{K}$).

Write the ideal gas law equation for a gas with constant pressure.

$P\Delta V=nR\Delta T$ (VI)

Here, $P$ is the pressure, and $\Delta V$ is the change in volume.

Solve equation (VI) for $\Delta V$ and expand the term $\Delta T$.

$\Delta V=\frac{nR\left(T_{\text{f}}-T_{\text{i}}\right)}{P}$ (VII)

Conclusion:

Substitute $1.00\text{mol}$ for $n$, $8.314\text{J/}\left(\text{mol}\cdot \text{K}\right)$ for $R$, $298.15\text{K}$ for $T_{\text{f}}$, $283.15\text{K}$ for $T_{\text{i}}$, and $1.00\text{atm}$ for $P$ in equation (VII) to find $\Delta V$.

$\begin{array}{c}\Delta V=\frac{\left(1.00\text{mol}\right)\left[8.314\text{J/}\left(\text{mol}\cdot \text{K}\right)\right]\left(298.15\text{K}-283.15{\text{K}}_{\text{i}}\right)}{1.00\text{atm}}\\ =\frac{\left(1.00\text{mol}\right)\left[8.314\text{J/}\left(\text{mol}\cdot \text{K}\right)\right]\left(298.15\text{K}-283.15{\text{K}}_{\text{i}}\right)}{1.00\text{atm}\times \frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}}\times \frac{1000\text{L}}{1{\text{m}}^3}\\ =1.23\text{L}\end{array}$

Therefore, the change in volume of the gas in the process is $\underset{\_}{1.23\text{L}}$.

(c)

To determine

The work done by the gas during the expansion.

Answer to Problem 18P

The work done by the gas during the expansion is $\underset{\_}{125\text{J}}$.

Explanation of Solution

Given that the pressure is $1.00\text{atm}$. It is obtained that the change in volume is $1.23\text{L}$.

Write the expression for the work done by a gas during expansion.

$W=P\Delta V$ (VIII)

Here, $W$ is the work done.

Conclusion:

Substitute $1.00\text{atm}$ for $P$, and $1.23\text{L}$ for $\Delta V$ in equation (VIII) to find $W$.

$\begin{array}{c}W=\left(1.00\text{atm}\right)\left(1.23\text{L}\right)\\ =\left(1.00\text{atm}\times \frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\left(1.23\text{L}\times \frac{1{\text{m}}^3}{1000\text{L}}\right)\\ =124.7\text{J}\\ \approx 125\text{J}\end{array}$

Therefore, the work done by the gas during the expansion is $\underset{\_}{125\text{J}}$.

(d)

To determine

The change in internal energy of the gas.

Answer to Problem 18P

The change in internal energy of the gas is $\underset{\_}{312\text{J}}$.

Explanation of Solution

It is obtained that the heat energy absorbed by the gas is $436.4\text{J}$, and the work done by the gas is $124.7\text{J}$.

Write the expression for change in internal energy according to first law of thermodynamics.

$\Delta U=Q-W$ (IX)

Here, $\Delta U$ is the change in internal energy.

Conclusion:

Substitute $436.4\text{J}$ for $Q$, and $124.7\text{J}$ for $W$ in equation (IX) to find $\Delta U$.

$\begin{array}{c}\Delta U=436.4\text{J}-124.7\text{J}\\ =312\text{J}\end{array}$

Therefore, the change in internal energy of the gas is $\underset{\_}{312\text{J}}$.