Problem 7 - Chemistry and Chemical Reactivity

Question

You have 3.5 L. of NO at a temperature of \(22.0^{\circ} \mathrm{C}\). What volume would the NO occupy at \(37^{\circ} \mathrm{C} ?\) (Assume the pressure is constant.)

Step-by-Step Solution

Answer

The volume at 37°C is approximately 3.68 L.

Step 1: Identify the Known Variables

We have the initial volume of NO, which is \(V_1 = 3.5 \, \text{L}\), and the initial temperature, which is \(T_1 = 22.0^{\circ}\text{C}\). We convert the initial temperature to Kelvin by adding 273, so \(T_1 = 22 + 273 = 295\, \text{K}\). The final temperature is \(T_2 = 37^{\circ}\text{C}\), so in Kelvin it is \(T_2 = 37 + 273 = 310\, \text{K}\). The pressure is constant in this scenario.

Step 2: Apply Charles's Law Formula

Charles's Law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) when the pressure is constant. Rearranging the formula to solve for the final volume \(V_2\), we get \(V_2 = V_1 \times \frac{T_2}{T_1}\).

Step 3: Plug in the Values

Substitute the known values into the formula: \(V_2 = 3.5 \, \text{L} \times \frac{310 \text{K}}{295 \text{K}}\).

Step 4: Calculate the Final Volume

Calculate the result to find \(V_2\):\[ V_2 = 3.5 \, \text{L} \times \frac{310}{295} \approx 3.68 \, \text{L} \]

Key Concepts

Temperature Conversion

When working with gas laws, it's crucial to use the correct temperature scale. In these problems, often presented in degrees Celsius ( T^{ ext{C}} ), we must convert to Kelvin ( T^{ ext{K}} ) for calculations. This is because Kelvin provides a true absolute scale required by the gas laws.
To convert from Celsius to Kelvin, use the straightforward formula: T^{ ext{K}} = T^{ ext{C}} + 273 . For example, if our initial temperature ( T_1 ) is 22.0°C, we add 273 to get 295 K. Similarly, for a temperature of 37°C, it becomes 310 K.
Remember, the Kelvin scale ensures we never have negative values in our gas laws. This is key for equations like Charles's Law, where temperature is directly related to volume.

Volume Calculation

Volume is a key aspect of gas problems, and you often need to calculate changes using specific laws. With Charles's Law, we focus on the relationship between temperature and volume, holding pressure constant. This law is excellent to determine how much space one mole of a gas will occupy at different temperatures.
Charles's Law formula is given by:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]Here, V_1 is the initial volume and T_1 is the initial temperature in Kelvin. V_2 is the final volume, and T_2 is the final temperature in Kelvin. By rearranging, we can easily solve for V_2:\[ V_2 = V_1 \times \frac{T_2}{T_1} \]Through this equation, determining the new volume is straightforward. For instance, in our exercise, starting with V_1 = 3.5 ext{ L} and temperatures T_1 = 295 ext{ K}, T_2 = 310 ext{ K}, plugging in these numbers into the equation gives us approximately V_2 = 3.68 ext{ L}.

Ideal Gas Law

The Ideal Gas Law is the comprehensive equation connecting pressure (P), volume (V), temperature (T in Kelvin), and the number of moles (n) of a gas. It's usually expressed as:\[ PV = nRT \]Where R is the ideal gas constant. While Charles's Law only examines the relationship between temperature and volume, the Ideal Gas Law integrates all essential aspects of a gas sample's conditions.
In Charles's Law problems like ours, the pressure is constant, and we're not dealing with specific moles or pressures directly, so we focus on the relationship between V and T: essentially a slice of the broader Ideal Gas Law picture.
Understanding the Ideal Gas Law helps see why temperature must be in Kelvin and provides a deeper appreciation of the interconnected nature of gases under varied conditions.