Question

Complete the following table for an ideal gas. $$\begin{aligned} &-\\\ &\begin{array}{|llll|} \hline {P} & {V} & {n} & {\boldsymbol{T}} \\ \hline 7.74 \times 10^{3} \mathrm{Pa} & 12.2 \mathrm{mL} & & 25^{\circ} \mathrm{C} \\ \hline & 43.0 \mathrm{mL} & 0.421 \mathrm{mol} & 223 \mathrm{K} \\ \hline 455 \text { torr } & & 4.4 \times 10^{-2} \mathrm{mol} & 331^{\circ} \mathrm{C} \\ \hline 745 \mathrm{mm} \mathrm{Hg} & 11.2 \mathrm{L} & 0.401 \mathrm{mol} & \\ \hline \end{array} \end{aligned}$$

Short answer

The completed table for the ideal gas is given by: $$ \begin{array}{|llll|} \hline P & V & n & T \\ \hline 7.74 \times 10^3\,\text{Pa} & 12.2\,\text{mL} & 0.00371\,\text{mol} & 25°C \\ \hline 2.1736\times10^5\,\text{Pa} & 43.0\,\text{mL} & 0.421\,\text{mol} & 223\,\text{K} \\ \hline 455\,\text{torr} & 3.631\,\text{mL} & 4.4\times10^{-2}\,\text{mol} & 331°C \\ \hline 745\,\text{mmHg} & 11.2\,\text{L}& 0.401\,\text{mol} & 332.45\,K \\ \hline \end{array} $$

Step-by-step solution

In the first row, we are given \(P = 7.74\times10^3Pa\), \(V = 12.2\,mL\), and \(T = 25°C\). We need to find n. First, we convert volume from mL to L: \[V = 12.2\,mL \cdot \frac{1\,L}{1000\,mL} = 0.0122\,L\] Then, we convert the temperature from Celsius to Kelvin: \[T = 25°C + 273.15 = 298.15\,K\] Now we can use the Ideal Gas Law to find n. Rearranging the equation to solve for n: \[n = \frac{PV}{RT}\] Plugging in the given values and the SI value of R: \[n = \frac{(7.74\times10^3\,Pa)(0.0122\,L)}{(8.314\,J/(mol\cdot K))(298.15\,K)}\] \[n \approx 0.00371\,mol\] #Step 2: Calculate "P" for the second row#

We have \(V = 43.0\,mL\), \(n = 0.421\,mol\), and \(T = 223\,K\). We are going to find P. Before we do so, we need to convert volume to L again: \[V = 43.0\,mL\cdot\frac{1\,L}{1000\,mL} = 0.043\,L\] Using the Ideal Gas Law and solving for P, we plug in the values: \[P = \frac{nRT}{V}\] \[P = \frac{(0.421\,mol)(8.314\,J/(mol\cdot K))(223\,K)}{0.043\,L}\] \[P \approx 2.1736\times10^5\,Pa\] #Step 3: Calculate "V" for the third row#

We have \(P = 455\,torr\), \(n = 4.4\times10^{-2}mol\), and \(T = 331°C\). We need to find V. Converting the pressure to Pa: \[P = 455\,torr \times \frac{101325\,Pa}{760\,torr} \approx 6.085\times10^4\,Pa\] And converting the temperature from Celsius to Kelvin: \[T = 331°C + 273.15 = 604.15\,K\] Now we can use the Ideal Gas Law to find V, rearranging the equation and plugging in the values: \[V = \frac{nRT}{P}\] \[V = \frac{(4.4\times10^{-2}\,mol)(8.314\,J/(mol\cdot K))(604.15\,K)}{6.085\times10^4\,Pa}\] \[V \approx 3.631\times10^{-3}\,L \;or\; 3.631\,mL\] #Step 4: Calculate "T" for the fourth row#

We are given \(P = 745\,mmHg\), \(V = 11.2\,L\), and \(n = 0.401\,mol\). We need to find T. Converting the pressure to Pa: \[P = 745\,mmHg\times\frac{101325\,Pa}{760\,mmHg} \approx 9.978\times10^4\,Pa\] Using the Ideal Gas Law to find T, rearranging the equation and plugging in the values: \[T = \frac{PV}{nR}\] \[T = \frac{(9.978\times10^4\,Pa)(11.2\,L)}{(0.401\,mol)(8.314\,J/(mol\cdot K))}\] \[T \approx 332.45\,K\] Now we've completed the whole table, and it can be represented like this: $$ \begin{array}{|llll|} \hline P & V & n & T \\ \hline 7.74 \times 10^3\,\text{Pa} & 12.2\,\text{mL} & 0.00371\,\text{mol} & 25°C \\ \hline 2.1736\times10^5\,\text{Pa} & 43.0\,\text{mL} & 0.421\,\text{mol} & 223\,\text{K} \\ \hline 455\,\text{torr} & 3.631\,\text{mL} & 4.4\times10^{-2}\,\text{mol} & 331°C \\ \hline 745\,\text{mmHg} & 11.2\,\text{L}& 0.401\,\text{mol} & 332.45\,K \\ \hline \end{array} $$

Key concepts

Gas Laws

The behavior of gases under various conditions of pressure, volume, and temperature is described by a set of laws known as gas laws. These laws are crucial for understanding how gases will react when subject to changes in their environment. One of the most fundamental of these is the Ideal Gas Law, which provides a simple equation to describe the state of an ideal gas. This law combines several other gas laws, including Boyle's Law (pressure inversely varies with volume at constant temperature), Charles's Law (volume directly varies with temperature at constant pressure), and Avogadro's Law (volume is directly proportional to the amount of gas at constant temperature and pressure).

Represented by the formula PV = nRT, where P stands for pressure, V for volume, n for the amount of substance (moles), R for the universal gas constant, and T for temperature in Kelvin, the Ideal Gas Law provides a reliable approximation for the behavior of real gases under many conditions, although it can lose accuracy under high pressure or low temperature where real gases deviate from ideal behavior.

Pressure Volume and Temperature Relationships

Understanding the pressure, volume, and temperature relationships of gases is key for many scientific and engineering calculations. According to the Ideal Gas Law, these variables are interrelated, and a change in any one of them can affect the others while keeping the amount of gas constant. To calculate these changes accurately, units need to be consistent, meaning pressures are often converted to pascals (Pa) from other units like torr or mmHg, and volumes usually from milliliters (mL) to liters (L) for standardization. Additionally, temperature must always be in Kelvin for gas law calculations as Celsius does not account for the absolute zero point where gases theoretically exert no pressure.

These relationships are used to solve real-world problems from predicting the expansion of gases in hot environments to finding the necessary pressure to store a certain gas volume. In practice, this can be demonstrated by manipulating the Ideal Gas Law to isolate any variable, based on the known quantities, to find the unknown variable.

Molar Volume of Gases

The molar volume of a gas is the volume occupied by one mole of the gas under specified conditions of temperature and pressure. For an ideal gas at standard temperature and pressure (0°C and 1 atm), the molar volume is approximately 22.4 L/mol. This is a key concept in chemistry because it allows for the prediction of gas volumes based on the amount of substance and conditions.

Exploring the interplay between molar volume and the Ideal Gas Law elucidates how changing conditions affect the space that a gas occupies. For example, increasing the temperature while holding pressure constant will cause the molar volume to increase due to the increased kinetic energy of the gas particles. Conversely, increasing the pressure will decrease the molar volume, as the gas particles are compressed into a smaller space. In calculations, identifying the molar amount and the conditions of temperature and pressure helps in determining the final volume, as seen in the exercise solutions provided.