Problem 7
Question
You have 3.5 L of \(\mathrm{NO}\) 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
Verified Answer
The final volume of \( \mathrm{NO} \) at \( 37^{\circ} \mathrm{C} \) is approximately 3.678 liters.
1Step 1: Identify the Known Values
The initial volume of the gas is given as \( V_1 = 3.5 \) liters. The initial temperature \( T_1 \) is \( 22.0^{\circ} \mathrm{C} \). This needs to be converted to Kelvin: \( T_1 = 22.0 + 273.15 = 295.15 \) K. The final temperature \( T_2 \) is \( 37^{\circ} \mathrm{C} \), which is \( T_2 = 37 + 273.15 = 310.15 \) K. We are asked to find the final volume \( V_2 \).
2Step 2: Apply Charles's Law
Charles's Law states that \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) when the pressure is held constant. We can rearrange this equation to solve for \( V_2 \): \[ V_2 = V_1 \times \frac{T_2}{T_1} \]
3Step 3: Calculate the Final Volume
Substitute the known values into the equation: \[ V_2 = 3.5 \times \frac{310.15}{295.15} \] Calculate \( \frac{310.15}{295.15} \approx 1.0508 \). Multiply this by the initial volume: \[ V_2 = 3.5 \times 1.0508 \approx 3.678 \] liters.
Key Concepts
Gas LawsTemperature ConversionVolume CalculationConstant Pressure
Gas Laws
Charles's Law is one of the fundamental gas laws, describing how gases behave under constant pressure. It specifically addresses the relationship between the volume and temperature of a gas. According to Charles's Law, when you increase the temperature of a gas, its volume increases, provided the pressure remains unchanged. This is because gas molecules move faster at higher temperatures, requiring more space. The key formula in Charles's Law is:
- \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
Temperature Conversion
Temperature conversion is crucial in applying Charles's Law accurately. This is because gas laws require the use of absolute temperatures, specifically Kelvin. To convert Celsius to Kelvin, you simply add 273.15 to the Celsius temperature.
- For instance, a temperature of \(22.0^{\circ} \text{C}\) becomes \(22.0 + 273.15 = 295.15 \text{K}\).
- Similarly, \(37^{\circ} \text{C}\) is \(37 + 273.15 = 310.15 \text{K}\).
Volume Calculation
Calculating the volume of a gas under different temperatures using Charles's Law involves a straightforward process. Once temperatures are converted to Kelvin, you can use the law's formula to find the new volume \(V_2\):\[ V_2 = V_1 \times \frac{T_2}{T_1} \]In our example, the initial volume \(V_1\) is 3.5 liters, with a temperature \(T_1\) of 295.15 K. The final temperature \(T_2\) is 310.15 K. By substituting these values into the equation, we find:\[ V_2 = 3.5 \times \frac{310.15}{295.15} \approx 3.678 \text{ liters} \]It's crucial to perform careful calculations, ensuring every step uses consistent units and is rounded correctly for precision.
Constant Pressure
Understanding the concept of constant pressure in the context of Charles's Law is key. It means that as other variables change (like temperature and volume), the pressure exerted by the gas remains unchanged.
- When gas is heated at constant pressure, its particles move faster, causing the gas to expand and requiring a larger volume to contain it.
- If pressure was not constant, other calculations might have been needed to adjust for changes in pressure.
Other exercises in this chapter
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