Question

Given the following sets of values for three of the gas variables, calculate the unknown quantity. a. \(P=782.4 \mathrm{mm}\) Hg; \(V=? ; n=0.1021\) mol; \(T=\) \(26.2^{\circ} \mathrm{C}\) b. \(P=? \mathrm{mm} \mathrm{Hg} ; V=27.5 \mathrm{mL} ; n=0.007812 \mathrm{mol}\) \(T=16.6^{\circ} \mathrm{C}\) c. \(P=1.045\) atm; \(V=45.2 \mathrm{mL} ; n=0.002241 \mathrm{mol}\) \(T=?^{\circ} \mathrm{C}\)

Short answer

In summary, for the three cases, we have calculated the unknown gas variables as follows: a. The volume \(V = 2.446 \mathrm{L}\) b. The pressure \(P = 668.3 \mathrm{mmHg}\) c. The temperature \(T = 50.8^{\circ}\mathrm{C}\)

Step-by-step solution

Case a: Finding V

In case a, we have P, n, and T, and we need to find V. Let's convert P from mmHg to atm and T from Celsius to Kelvin before we apply the Ideal Gas Law. 1. Convert P: \(\frac{782.4 \mathrm{mmHg}}{760\mathrm{mmHg/atm}} = 1.0295 \mathrm{atm}\) 2. Convert T: \(26.2^{\circ}\mathrm{C} + 273.15 = 299.35 \mathrm{K}\) 3. Rearrange Ideal Gas Law to solve for V: \(V = \frac{nRT}{P}\) 4. Plug in values: \[\begin{aligned} V &= \frac{(0.1021 \mathrm{mol})(0.0821\frac{\mathrm{L\cdot atm}}{\mathrm{K\cdot mol}})(299.35\mathrm{K})}{1.0295\mathrm{atm}}\\ V &=2.446 \mathrm{L} \end{aligned}\] So, the volume \(V = 2.446 \mathrm{L}\).

Case b: Finding P

In case b, we have V, n, and T. We need to find P. Convert V from mL to L and T from Celsius to Kelvin before applying the Ideal Gas Law. 1. Convert V: \(\frac{27.5\mathrm{mL}}{1000\mathrm{mL/L}} = 0.0275\mathrm{L}\) 2. Convert T: \(16.6^{\circ}\mathrm{C} + 273.15 = 289.75\mathrm{K}\) 3. Rearrange Ideal Gas Law to solve for P: \(P = \frac{nRT}{V}\) 4. Plug in values: \[\begin{aligned} P &= \frac{(0.007812\mathrm{mol})(0.0821\frac{\mathrm{L\cdot atm}}{\mathrm{K\cdot mol}})(289.75\mathrm{K})}{0.0275\mathrm{L}}\\ P &=0.8793\mathrm{atm} \end{aligned}\] 5. Convert P from atm to mmHg: \(0.8793\mathrm{atm}\cdot 760\frac{\mathrm{mmHg}}{\mathrm{atm}} = 668.3\mathrm{mmHg}\) So, the pressure \(P = 668.3 \mathrm{mmHg}\).

Case c: Finding T

In case c, we have P, V, and n. We need to find T. Convert P from atm to atm (it's already in atm) and V from mL to L before applying the Ideal Gas Law. 1. Convert V: \(\frac{45.2\mathrm{mL}}{1000\mathrm{mL/L}} = 0.0452\mathrm{L}\) 2. Rearrange Ideal Gas law to solve for T: \(T = \frac{PV}{nR}\) 3. Plug in values: \[\begin{aligned} T &= \frac{(1.045\mathrm{atm})(0.0452\mathrm{L})}{(0.002241\mathrm{mol})(0.0821\frac{\mathrm{L\cdot atm}}{\mathrm{K\cdot mol}})}\\ T &=323.95\mathrm{K} \end{aligned}\] 4. Convert T from Kelvin to Celsius: \(323.95\mathrm{K} - 273.15 = 50.8^{\circ}\mathrm{C}\) So, the temperature \(T = 50.8^{\circ}\mathrm{C}\).

Key concepts

Understanding Gas Variables

Gas variables are the cornerstone of solving problems related to the behavior of gases. These variables include pressure (P), which is often measured in atmospheres (atm) or millimeters of mercury (mmHg), and is indicative of the force gas exerts on its container walls. Volume (V), typically in liters (L), defines the space the gas occupies. The amount of gas in moles (n) reveals the quantity of gas particles present. Finally, temperature (T), which is stated in Kelvin (K), relates to the average kinetic energy of the gas particles.

The Ideal Gas Law, expressed as PV = nRT, unifies these variables into a single equation. The constant \(R\) is the ideal gas constant, which bridges these units together. Working with this law involves manipulating these variables and ensures a deep understanding of gaseous behavior to solve for an unknown value.

mmHg to atm Conversion

Converting pressure from mmHg to atm is a critical step when working with the Ideal Gas Law, as it typically requires pressure in atm. The conversion factor is based on the equivalence of 760 mmHg to 1 atm, the standard atmospheric pressure. To convert, you divide the given pressure in mmHg by the conversion factor, 760 mmHg/atm.

For example, to convert 782.4 mmHg to atm:
\[\frac{782.4 \mathrm{mmHg}}{760 \mathrm{mmHg/atm}} = 1.0295 \mathrm{atm}\]
Understanding these units and being able to convert between them is vital as it guarantees accuracy when applying the Ideal Gas Law.

Celsius to Kelvin Conversion

Temperature conversions are often necessary in gas law problems because the Ideal Gas Law requires temperature in Kelvin. Kelvin is the SI unit for thermodynamic temperature and is an absolute scale, meaning it starts at absolute zero, or 0 K, which is the theoretical point where particles have no thermal energy.

To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. This is because 0 degrees Celsius is equivalent to 273.15 Kelvin, aligning the scales. For instance, if the temperature is 26.2°C:
\[26.2^\circ\mathrm{C} + 273.15 = 299.35 \mathrm{K}\]
This conversion is crucial to solving gas-related problems because it puts temperature in terms that reflect the kinetic energy per particle, without negative values, which are conceptually incompatible with the kinetic theory of gases.

Solving for Gas Law Variables

When one variable of the Ideal Gas Law needs to be determined, the first step is to ensure all other values are in the correct units - atm for pressure, liters for volume, moles for the amount of substance, and Kelvin for temperature. The next step is rearranging the equation algebraically to solve for the unknown value.

For example, if solving for volume (V), the Ideal Gas Law is rearranged as \(V = \frac{nRT}{P}\). By substituting the values of n, R, T, and P, and performing the multiplication and division, the volume can be calculated. The process is similar for other variables - isolate the unknown on one side of the equation and compute its value based on the given information and the ideal gas constant R.

As seen in the given exercises, recognizing the variable to solve for and correctly converting all the other values into their compatible units is paramount to revealing the unknown and thus understanding the property of the gas in the context of the situation.