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A sample of gas has a volume of \(2.50 \mathrm{~L}\) at a pressure of \(670 . \mathrm{mmHg}\) and a temperature of \(80 .{ }^{\circ} \mathrm{C} .\) If the pressure remains constant but the temperature is decreased, the gas occupies \(1.25 \mathrm{~L}\). Determine this new temperature, in degrees Celsius.

Short Answer

Expert verified
The new temperature is \(-97 \text{ }^{\circ}\text{C} \).

Step by step solution

01

Use the Combined Gas Law

The problem states that the pressure is constant, so we can use the formula connecting temperature and volume: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] where \( V_1 \) is the initial volume, \( T_1 \) is the initial temperature, \( V_2 \) is the final volume, and \( T_2 \) is the final temperature. Since the temperatures are in Celsius, we need to convert them to Kelvin.
02

Convert Initial Temperature to Kelvin

To convert the initial temperature from Celsius to Kelvin, use the formula: \[ T(K) = T(^{\circ}C) + 273.15 \] Therefore, \[ T_1 = 80 + 273.15 = 353.15 \text{ K} \].
03

Write Down Known Values

The known values are:\[ V_1 = 2.50 \text{ L} \] \[ T_1 = 353.15 \text{ K} \] \[ V_2 = 1.25 \text{ L} \] We are solving for \( T_2 \).
04

Solve for Final Temperature \( T_2 \) in Kelvin

Rearrange the combined gas law equation to solve for \( T_2 \): \[ T_2 = \frac{V_2 \cdot T_1}{V_1} \] Substitute the known values into the equation: \[ T_2 = \frac{1.25 \cdot 353.15}{2.50} = 176.575 \text{ K} \].
05

Convert \( T_2 \) to Degrees Celsius

Convert the final temperature from Kelvin back to degrees Celsius: \[ T(^{\circ}C) = T(K) - 273.15 \] Applying the values, \[ T_2 = 176.575 - 273.15 = -96.575 \text{ }^{\circ}\text{C} \].
06

Round to Appropriate Significant Figures

The final result should be rounded according to the significant figures in the given data, which is two significant figures based on the volumes and temperatures given. Thus, the temperature \( T_2 \) is \(-97 \text{ }^{\circ}\text{C} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume and Temperature Relationship
When dealing with gases, understanding how volume and temperature interact is crucial. **The Combined Gas Law** provides a straightforward way to relate these properties while holding pressure constant. It's expressed as: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] where:
  • \( V_1 \) and \( V_2 \) are the initial and final volumes
  • \( T_1 \) and \( T_2 \) are the initial and final temperatures
A direct relationship exists here: When the temperature of a gas increases, its volume increases proportionally if the pressure remains constant, and vice versa. This means if you decrease the temperature, the volume decreases as well. Perfect for problems involving constant pressure and changing temperature and volume. Break your calculations into smaller parts, always ensuring the units are consistent.
Kelvin to Celsius Conversion
Temperature conversions between Kelvin and Celsius are straightforward and essential in gas law calculations. **The formula for converting Celsius to Kelvin** is: \[ T(K) = T(^{\circ}C) + 273.15 \]Every Celsius degree equals exactly one Kelvin, which simplifies conversion. When using gas laws, always work in Kelvin to ensure formula correctness. Celsius can produce misleading answers because it starts at an arbitrary point (freezing of water). For converting back from Kelvin to Celsius:
  • **Subtract 273.15** to switch your Kelvin result back to Celsius.
Remember, Kelvin is used for scientific calculations because it is an absolute scale, meaning it starts at absolute zero, where no thermal motion occurs.
Gas Law Calculations
Gas law calculations often require using mathematical formulas to relate temperature, volume, and other properties. For example, with **the Combined Gas Law**, once you've set your known values:
  • Identify what needs to be calculated (e.g., new temperature or volume).
  • Rearrange the formula to solve for the unknown variable. For temperature: \[ T_2 = \frac{V_2 \cdot T_1}{V_1} \]
  • Insert your known values and solve the equation.
Always check the units. Volumes should line up (both in liters or milliliters), and temperatures should be in Kelvin before substituting. After calculations, verify results by converting units back if needed, such as changing Kelvin temperatures to Celsius for practical interpretation. Rounding off final answers is crucial – follow significant figure rules for precision, often dictated by initial conditions in the problem. An excellent check ensures the result's plausibility within the context of the problem.

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Most popular questions from this chapter

Consider a sample of \(\mathrm{N}_{2}\) gas under conditions in which it obeys the ideal gas law exactly. Which of these statements is/are true? (a) A sample of \(\mathrm{Ne}(\mathrm{g})\) under the same conditions must obey the ideal gas law exactly. (b) The speed at which one particular \(\mathrm{N}_{2}\) molecule is moving changes from time to time. (c) Some \(\mathrm{N}_{2}\) molecules are moving more slowly than some of the molecules in a sample of \(\mathrm{O}_{2}(\mathrm{~g})\) under the same conditions. (d) Some \(\mathrm{N}_{2}\) molecules are moving more slowly than some of the molecules in a sample of \(\mathrm{Ne}(\mathrm{g})\) under the same conditions. (e) When two \(\mathrm{N}_{2}\) molecules collide, it is possible that both may be moving faster after the collision than they were before.

Use the postulates of the kinetic-molecular theory to explain each phenomenon. (a) \(\mathrm{Br}_{2}(\mathrm{~g})\) is reddish brown and transparent; \(\mathrm{Br}_{2}(\ell)\) is very dark brown and very little light passes through it. (b) When equal volumes of \(\mathrm{Br}_{2}(\mathrm{~g})\) and \(\mathrm{N}_{2}(\mathrm{~g})\) at the same \(T\) and \(P\) are brought into contact, they mix rapidly and the color is only half as dark as the initial \(\mathrm{Br}_{2}\) color.

Arrange these four gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{Cl}_{2}, \mathrm{~F}_{2}, \mathrm{~N}_{2},\) and \(\mathrm{O}_{2}\).

Formaldehyde, \(\mathrm{CH}_{2} \mathrm{O},\) is a volatile organic compound that is sometimes released from insulation used in home construction, and it can be trapped and build up inside the home. When this happens, people exposed to the formaldehyde can suffer adverse health effects. The U. S. National Institute of Occupational Health and Safety (NIOSH) guideline for the maximum allowable concentration of formaldehyde in air in the workplace is \(16 \mathrm{ppb}\) (parts per billion) for an eight-hour average exposure. (a) Determine the partial pressure of formaldehyde at the maximum allowable level of \(16 \mathrm{ppb}\). (b) Calculate how many molecules of formaldehyde are present in each cubic centimeter of air when formaldehyde is present at \(16 \mathrm{ppb}\). (c) Calculate how many total molecules of formaldehyde are present in a room: \(15.0 \mathrm{ft}\) long \(\times 10.0 \mathrm{ft}\) wide \(X\) \(8.00 \mathrm{ft}\) high (at \(16 \mathrm{ppb}\) ).

Calculate the densities of \(\mathrm{Cl}_{2}\) and of \(\mathrm{SO}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(0.750 \mathrm{~atm} .\) Then, calculate the density of \(\mathrm{Cl}_{2}\) at \(35^{\circ} \mathrm{C}\) and \(0.750 \mathrm{~atm}\) and the density of \(\mathrm{SO}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(2.60 \mathrm{~atm} .\)

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