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A mass of ideal gas occupies \(38 \mathrm{~mL}\) at \(20^{\circ} \mathrm{C}\). If its pressure is held constant, what volume does it occupy at a temperature of \(45^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The volume of the gas at 45°C is approximately 41.2 mL.

Step by step solution

01

Understanding Charles's Law

Charles's Law relates the volume and temperature of a gas at constant pressure. The formula is given by \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( V_1 \) and \( T_1 \) are the initial volume and temperature, and \( V_2 \) and \( T_2 \) are the final volume and temperature of the gas.
02

Converting Temperatures to Kelvin

To use Charles's Law, convert the temperatures from Celsius to Kelvin. The initial temperature, \( T_1 \), is \( 20^{\circ} \mathrm{C} = 20 + 273.15 = 293.15 \, \mathrm{K} \). The final temperature, \( T_2 \), is \( 45^{\circ} \mathrm{C} = 45 + 273.15 = 318.15 \, \mathrm{K} \).
03

Setting Up the Equation

Substitute the given values into Charles's Law equation. Here, \( V_1 = 38 \, \mathrm{mL} \), \( T_1 = 293.15 \, \mathrm{K} \), and \( T_2 = 318.15 \, \mathrm{K} \). The equation becomes \( \frac{38}{293.15} = \frac{V_2}{318.15} \).
04

Solving for Final Volume

Rearrange the equation to solve for \( V_2 \):\[ V_2 = \frac{38 \times 318.15}{293.15} \]Calculate \( V_2 \) to find the final volume.
05

Calculating the Answer

Perform the calculation: \[ V_2 = \frac{38 \times 318.15}{293.15} \approx 41.2 \, \mathrm{mL} \]The final volume the gas occupies at \( 45^{\circ} \mathrm{C} \) is approximately \( 41.2 \, \mathrm{mL} \).

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry and physics that relates the pressure, volume, temperature, and amount of gas. It is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the absolute temperature measured in Kelvin. This equation provides a comprehensive description of gas behavior under various conditions.
The ideal gas law combines several simpler laws, including Charles's Law, Boyle's Law, and Avogadro's Law, offering a complete view of how gases behave when one variable changes while others remain constant. Although real gases deviate from ideal behavior at high pressures and low temperatures, this law is remarkably accurate for many gases under a wide range of conditions.
Volume and Temperature Relationship
The relationship between volume and temperature of a gas at constant pressure is described by Charles's Law. This law states that the volume of an ideal gas is directly proportional to its absolute temperature if the pressure remains unchanged. Mathematically, this is expressed as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).
Charles's Law explains that when a gas’s temperature increases, its volume increases as well, provided its pressure remains the same. Conversely, if the temperature decreases, the volume will also decrease. This principle is key in understanding how gases expand and contract with temperature changes, such as a hot air balloon rising and falling in response to the temperature of the air inside.
Temperature Conversion to Kelvin
Temperature conversion is essential for solving problems involving gas laws, as these laws require temperature to be in Kelvin. Celsius and Kelvin are both units for measuring temperature, but Kelvin is the standard used in scientific calculations because it is an absolute temperature scale.
  • To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature. For example, \( T_{\mathrm{Celsius}} = 20^{\circ} \mathrm{C} \) can be converted to \( T_{\mathrm{Kelvin}} = 20 + 273.15 = 293.15 \mathrm{K} \).
  • This conversion ensures that all scientific calculations are consistent and align with the absolute scale, where 0 Kelvin represents absolute zero, the theoretical point at which particles have no thermal energy.
Ensuring temperature is in Kelvin avoids negative values and maintains the direct proportionality required by Charles's Law.
Gas Laws in Physics
Gas laws are the bedrock of understanding how gases behave under different conditions. Numerous laws describe the relationships between the variables of gases, such as pressure, volume, and temperature. Some of the key gas laws include:
  • **Boyle's Law:** Describes how pressure and volume are inversely proportional when temperature is held constant.
  • **Charles's Law:** Demonstrates that volume is directly proportional to temperature when pressure remains fixed.
  • **Avogadro's Law:** Establishes that volume is directly proportional to the number of moles of the gas at constant temperature and pressure.
These laws are primarily applicable to ideal gases but often provide accurate approximations for real gases under a range of regular conditions.
Learning these laws aids students in predicting how gas reacts to changes in conditions, and they form the foundation for more advanced topics in thermodynamics and fluid mechanics.

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Most popular questions from this chapter

A gas is sealed into a closed container. The gas is heated so that its temperature rises from \(100^{\circ} \mathrm{C}\) to \(200{ }^{\circ} \mathrm{C}\). If the initial absolute pressure in the container was \(2.00\) atm, what will its new value be?

A cylindrical diving bell (a vertical cylinder with open bottom end and closed top end) \(12.0 \mathrm{~m}\) high is lowered in a lake until water within the bell rises \(8.0 \mathrm{~m}\) from the bottom end. Determine the distance from the top of the bell to the surface of the lake. (Atmospheric pressure \(=1.00\) atm. )

A mass of oxygen occupies \(0.0200 \mathrm{~m}^{3}\) at atmospheric pressure, \(101 \mathrm{kPa}\), and \(5.0{ }^{\circ} \mathrm{C}\). Determine its volume if its pressure is increased to \(108 \mathrm{kPa}\) while its temperature is changed to \(30^{\circ} \mathrm{C}\). From $$ \frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \quad \text { we have } \quad V_{2}=V_{1}\left(\frac{P_{1}}{P_{2}}\right)\left(\frac{T_{2}}{T_{1}}\right) $$ But \(T_{1}=5+273=278 \mathrm{~K}\) and \(T_{2}=30+273=303 \mathrm{~K}\); consequently, $$ V_{2}=\left(0.0200 \mathrm{~m}^{3}\right)\left(\frac{101}{108}\right)\left(\frac{303}{278}\right)=0.0204 \mathrm{~m}^{3} $$

A \(5000-\mathrm{cm}^{3}\) tank contains an ideal gas \((M=40 \mathrm{~kg} / \mathrm{kmol})\) at a gauge pressure of \(530 \mathrm{kPa}\) and a temperature of \(25^{\circ} \mathrm{C}\). Assuming atmospheric pressure to be \(100 \mathrm{kPa}\), what mass of gas is in the tank?

A sealed tank having a volume of \(25.0 \times 10^{-3} \mathrm{~m}^{3}\) contains \(0.56 \mathrm{~kg}\) of nitrogen \(\left(\mathrm{N}_{2}\right)\). How many kilomoles of gas are in the tank? [Hint: The atomic mass of the molecule is 28.]

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