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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

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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.

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
The Combined Gas Law shows the inverse relationship between pressure and volume while indicating the direct relationship between volume and 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
Understanding ideal gas behavior simplifies problem-solving and lets us use equations like the Combined Gas Law more effectively.
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
For instance, in our initial problem, knowing how to convert temperature and pressure correctly and applying the Combined Gas Law allowed us to find the volume of COâ‚‚ at new conditions. Remember to always double-check units and calculations for accuracy to avoid errors.

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Most popular questions from this chapter

Your spaceship has docked at a space station above Mars. The temperature inside the space station is a carefully controlled \(24{ }^{\circ} \mathrm{C}\) at a pressure of \(745 \mathrm{mmHg}\). A balloon with a volume of \(425 \mathrm{~mL}\) drifts into the airlock where the temperature is \(-95^{\circ} \mathrm{C}\) and the pressure is \(0.115 \mathrm{~atm}\). What is the new volume of the balloon \((n\) remains constant)? Assume that the balloon is very elastic.

A gas mixture containing oxygen, nitrogen, and neon exerts a total pressure of \(1.20 \mathrm{~atm} .\) If helium added to the mixture increases the pressure to \(1.50 \mathrm{~atm}\), what is the partial pressure (atm) of the helium?

In a gas mixture, the partial pressures are argon \(415 \mathrm{mmHg}\), neon \(75 \mathrm{mmHg}\), and nitrogen \(125 \mathrm{mmHg}\). What is the total pressure (atm) exerted by the gas mixture?

A sample of hydrogen \(\left(\mathrm{H}_{2}\right)\) gas at \(127{ }^{\circ} \mathrm{C}\) has a pressure of \(2.00 \mathrm{~atm}\). At what temperature \(\left({ }^{\circ} \mathrm{C}\right)\) will the pressure of the \(\mathrm{H}_{2}\) decrease to \(0.25 \mathrm{~atm}\), if \(V\) and \(n\) are constant?

Two flasks of equal volume and at the same temperature contain different gases. One flask contains \(1.00 \mathrm{~g}\) of \(\mathrm{Ne}\), and the other flask contains \(1.00 \mathrm{~g}\) of He. Which of the following statements are correct? Explain your answers. a. Both flasks contain the same number of atoms. b. The pressures in the flasks are the same. c. The flask that contains helium has a higher pressure than the flask that contains neon. d. The densities of the gases are the same.

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