Ideal Gas Law Calculation Pressure Of A Gas Sample

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In the realm of chemistry, understanding the behavior of gases is paramount. The ideal gas law, a cornerstone of chemical principles, provides a fundamental equation to describe the relationship between pressure, volume, temperature, and the number of moles of a gas. This article delves into the application of the ideal gas law to determine the pressure of a gas under specific conditions. We will walk through a step-by-step calculation, ensuring clarity and comprehension of this vital concept. Our focus will be on a sample problem where we need to find the pressure of a gas given its amount in moles, volume, and temperature. This exercise will not only reinforce the theoretical understanding of the ideal gas law but also demonstrate its practical application in solving real-world problems.

Let's consider a scenario where a sample of gas contains 6.25 x 10^-3 mol and is confined within a 500.0 mL flask at a temperature of 265°C. Our objective is to determine the pressure exerted by this gas, expressed in kilopascals (kPa). This problem is a classic application of the ideal gas law, and by solving it, we will illustrate how this law connects the macroscopic properties of a gas.

At the heart of this calculation lies the ideal gas law, a simple yet powerful equation:

PV=nRTPV = nRT

Where:

  • P represents the pressure of the gas.
  • V signifies the volume occupied by the gas.
  • n denotes the number of moles of the gas.
  • R is the ideal gas constant.
  • T stands for the absolute temperature of the gas.

The ideal gas law is a fundamental principle in chemistry that describes the state of a gas under ideal conditions. It is an equation of state that relates the pressure, volume, temperature, and number of moles of a gas. The ideal gas law assumes that gas molecules have negligible volume and do not interact with each other. While no real gas is perfectly ideal, the ideal gas law provides a good approximation for the behavior of many gases under a wide range of conditions. The equation is expressed as PV = nRT, where P is the pressure of the gas, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Understanding and applying the ideal gas law is crucial for solving various problems in chemistry, such as determining gas densities, molar masses, and the stoichiometry of reactions involving gases. In this article, we will use the ideal gas law to calculate the pressure of a gas given its volume, temperature, and number of moles.

Before we dive into the calculations, let's identify the variables provided in the problem statement:

  • Number of moles (n) = 6.25 x 10^-3 mol
  • Volume (V) = 500.0 mL
  • Temperature (T) = 265°C

The ideal gas law, expressed as PV = nRT, is a fundamental equation in chemistry that describes the state of a gas under ideal conditions. To effectively use this law, it's crucial to correctly identify the given variables in a problem. In our specific problem, we are given the number of moles (n) of the gas, which is 6.25 x 10^-3 mol. This value tells us the amount of gas present in the sample. We are also given the volume (V) that the gas occupies, which is 500.0 mL. The volume is a measure of the space the gas fills. Lastly, we have the temperature (T) of the gas, which is 265°C. Temperature is a critical factor in gas behavior, as it affects the kinetic energy of the gas molecules. Correctly identifying these variables is the first step in applying the ideal gas law to solve for the unknown variable, which in this case is the pressure (P) of the gas. This careful approach ensures that we can proceed with the calculation accurately and efficiently. Therefore, understanding the given variables is not just about noting the numbers but also about grasping their physical significance in the context of the problem.

To ensure compatibility with the ideal gas constant (R = 8.31 L⋅kPa/mol⋅K), we need to convert the given volume from milliliters (mL) to liters (L) and the temperature from Celsius (°C) to Kelvin (K).

  • Volume conversion:

    • 500.0 mL = 500.0 / 1000 L = 0.5000 L

  • Temperature conversion:

    • T(K) = T(°C) + 273.15
    • T(K) = 265°C + 273.15 = 538.15 K

Unit conversions are a critical step in solving problems involving the ideal gas law. The ideal gas constant, R, is typically given in units of L⋅kPa/mol⋅K, which means that the volume, pressure, and temperature must be in liters, kilopascals, and Kelvin, respectively. Our problem provides the volume in milliliters (mL) and the temperature in Celsius (°C), necessitating conversion before we can apply the ideal gas law. Converting the volume from milliliters to liters involves dividing the volume in mL by 1000, as there are 1000 mL in 1 L. Thus, 500.0 mL is converted to 0.5000 L. For temperature, the conversion from Celsius to Kelvin is done by adding 273.15 to the Celsius temperature. Therefore, 265°C is converted to 538.15 K. These conversions are not just about changing the numerical value; they ensure that the units are consistent throughout the calculation. Using the correct units is essential for obtaining an accurate result when applying the ideal gas law. This step highlights the importance of attention to detail and a thorough understanding of unit relationships in scientific calculations. By performing these conversions accurately, we set the stage for a successful application of the ideal gas law and the determination of the gas pressure.

Now that we have all the variables in the correct units, we can rearrange the ideal gas law equation to solve for pressure (P):

P=nRTVP = \frac{nRT}{V}

Substituting the known values:

P=(6.25imes10−3mol)(8.31LimeskPa/molimesK)(538.15K)0.5000LP = \frac{(6.25 imes 10^{-3} mol)(8.31 L imes kPa / mol imes K)(538.15 K)}{0.5000 L}

Calculating the pressure:

P=27.940.5000kPaP = \frac{27.94}{0.5000} kPa

P=55.88kPaP = 55.88 kPa

Applying the ideal gas law to solve for pressure involves a straightforward algebraic manipulation and substitution of known values. The ideal gas law equation, PV = nRT, can be rearranged to solve for pressure (P) by dividing both sides of the equation by volume (V), resulting in P = nRT/V. This rearranged equation allows us to directly calculate the pressure when we know the number of moles (n), the ideal gas constant (R), the temperature (T), and the volume (V). In our problem, we have already identified and converted the necessary variables: n = 6.25 x 10^-3 mol, R = 8.31 Lâ‹…kPa/molâ‹…K, T = 538.15 K, and V = 0.5000 L. Substituting these values into the equation, we get P = (6.25 x 10^-3 mol * 8.31 Lâ‹…kPa/molâ‹…K * 538.15 K) / 0.5000 L. Performing the multiplication in the numerator gives us a value of approximately 27.94. Dividing this by the volume of 0.5000 L yields a pressure of 55.88 kPa. This calculation demonstrates the power and utility of the ideal gas law in quantitatively describing the behavior of gases. By carefully applying the equation and paying attention to units, we can accurately determine the pressure of a gas under given conditions. The result, 55.88 kPa, is the pressure exerted by the gas in the flask, calculated using the ideal gas law.

Therefore, the pressure of the gas in the 500.0 mL flask at 265°C is approximately 55.88 kPa. This result showcases the application of the ideal gas law in determining gas pressure under specific conditions. The ideal gas law, a cornerstone of chemistry, allows us to predict the behavior of gases based on fundamental relationships between pressure, volume, temperature, and the number of moles. By accurately identifying given variables, performing necessary unit conversions, and applying the ideal gas law equation, we can confidently solve for unknown parameters, such as pressure. This problem-solving approach not only reinforces our understanding of theoretical concepts but also equips us with practical skills applicable in various scientific and engineering contexts. The pressure of 55.88 kPa is a direct consequence of the gas molecules colliding with the walls of the flask, and it is influenced by the temperature and the amount of gas present. Understanding these relationships is crucial for anyone studying chemistry or related fields. This exercise underscores the importance of the ideal gas law as a tool for quantitatively analyzing and predicting gas behavior in a variety of applications.

  • Ideal Gas Law
  • Gas Pressure Calculation
  • Moles
  • Volume
  • Temperature
  • kPa
  • Unit Conversions
  • Chemistry
  • Gas Behavior
  • Problem Solving
  • Ideal Gas Constant
  • PV=nRT