Chapter 6: Problem 72
In the fermentation of glucose (wine making), \(780 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) gas was produced at \(37^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\) What is the volume (L) of the gas when measured at \(22{ }^{\circ} \mathrm{C}\) and \(675 \mathrm{mmHg}\) ?
Short Answer
Expert verified
0.835 L
Step by step solution
01
Identify the Known Variables
Before solving the problem, let's identify the known variables and convert them to appropriate units if needed. We have:Initial volume, \(V_1 = 780 \text{ mL} = 0.780 \text{ L}\)Initial temperature, \(T_1 = 37^{\text{o}}\text{C} = 37 + 273 = 310 \text{ K}\)Initial pressure, \(P_1 = 1.00 \text{ atm}\)Final temperature, \(T_2 = 22^{\text{o}}\text{C} = 22 + 273 = 295 \text{ K}\)Final pressure, \(P_2 = 675 \text{ mmHg} = \frac{675}{760} \text{ atm} = 0.888 \text{ atm}\)
02
Apply the Combined Gas Law
With the variables identified, use the Combined Gas Law:\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]Isolating \(V_2\):\[V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}\]
03
Substitute the Known Values
Substitute the known values into the equation found in Step 2:\[V_2 = \frac{(1.00 \text{ atm})(0.780 \text{ L})(295 \text{ K})}{(0.888 \text{ atm})(310 \text{ K})}\]
04
Calculate the Result
Perform the calculation:\[V_2 \approx \frac{(1.00)(0.780)(295)}{(0.888)(310)} \approx \frac{229.8}{275.28} \approx 0.835 \text{ L}\]
05
Conclusion
The volume of the gas at the new conditions is approximately 0.835 L.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gas Laws
Gas laws describe how the volume, pressure, and temperature of a gas relate to each other. They provide mathematical relationships that help predict how a gas will behave under different conditions. The Combined Gas Law is one such relationship, combining Boyle's, Charles's, and Gay-Lussac's laws into one equation: \[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\] where:
- \( P \) represents pressure
- \( V \) represents volume
- \( T \) represents temperature
Temperature and Pressure Conversions
Conversions ensure that the units for temperature and pressure are consistent when using gas laws. For temperature, you must convert degrees Celsius to Kelvin because gas law equations require absolute temperature. The conversion is simple: \[T_{\mathrm{K}} = T_{\mathrm{C}} + 273\] For pressure, units can vary (atm, mmHg, kPa, etc.), so it's essential to convert to the same unit. If you're given pressure in mmHg and need to convert to atm, use the following conversion: \[ 1 \: \mathrm{atm} = 760 \: \mathrm{mmHg} \] So, to convert from mmHg to atm: \[ P_{\mathrm{atm}} = \frac{P_{\mathrm{mmHg}}}{760}\] These conversions are crucial for accurate calculations in gas law problems.
Ideal Gas Behavior
Ideal gas behavior assumes that gas molecules do not attract or repel each other and occupy negligible volume. Real gases deviate from ideal behavior at high pressures and low temperatures. However, for most gas law calculations, we assume ideal gas behavior for simplicity. Using the ideal gas law, we can derive approximate solutions for real-world gas problems: \[PV = nRT\] where:
- \( P \) is pressure
- \( V \) is volume
- \( n \) is the number of moles of gas
- \( R \) is the universal gas constant
- \( T \) is temperature in Kelvin
Problem-Solving in Chemistry
Effective problem-solving in chemistry involves understanding the core principles and applying them step by step. When tackling gas law problems:
- Identify all known variables and convert to appropriate units
- Choose the correct gas law based on the given data
- Manipulate the equation to solve for the unknown variable
- Substitute the known values into the equation
- Perform the calculation carefully