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Chapter 5: Problem 41

A gas at \(772 \mathrm{mmHg}\) and \(35.0^{\circ} \mathrm{C}\) occupies a volume of 6.85 L. Calculate its volume at STP.

Short Answer

Expert verified
The volume at STP is 6.48 L.

Step by step solution

01

Understand the Problem

We need to find the new volume of a gas when it is moved to standard temperature and pressure (STP) conditions. STP conditions are defined as a temperature of 0掳C (273 K) and a pressure of 1 atm (760 mmHg). We will use the Combined Gas Law to solve this.
02

Write the Combined Gas Law Formula

The Combined Gas Law is given by: \(\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}\), where \(P\) is pressure, \(V\) is volume, and \(T\) is temperature.
03

Convert Initial Temperature to Kelvin

The initial temperature is given as 35.0掳C. We convert this to Kelvin using the formula \(T(K) = T(掳C) + 273\). Thus, \(T_1 = 35.0 + 273 = 308\, K\).
04

Identify Known and Unknown Variables

We are given: \(P_1 = 772\, \mathrm{mmHg}\), \(V_1 = 6.85\, \mathrm{L}\), \(T_1 = 308\, \mathrm{K}\). The STP conditions are \(P_2 = 760\, \mathrm{mmHg}\) and \(T_2 = 273\, \mathrm{K}\). We need to find \(V_2\).
05

Substitute Values into the Combined Gas Law

Substitute the known values into the Combined Gas Law: \(\frac{772 \times 6.85}{308} = \frac{760 \times V_2}{273}\).
06

Solve for \(V_2\)

Rearrange the equation to solve for \(V_2\): \(V_2 = \frac{772 \times 6.85 \times 273}{308 \times 760}\).
07

Calculate \(V_2\)

Perform the calculation: \(V_2 = \frac{772 \times 6.85 \times 273}{308 \times 760} = 6.48\, \mathrm{L}\).

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Key Concepts

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

STP (Standard Temperature and Pressure)
Understanding the concept of Standard Temperature and Pressure (STP) is crucial when dealing with gases and their behaviors under different conditions. **STP** refers to a set of standard conditions for measuring gas volumes and is used as a reference to ensure consistency across scientific studies.
  • **Temperature at STP**: The temperature is defined as 0 degrees Celsius, which equals 273 Kelvin. Kelvin is the unit of measurement used in gas law calculations as it keeps temperatures positive and simplifies equations.
  • **Pressure at STP**: The standard pressure is 1 atmosphere (atm), or 760 millimeters of mercury (mmHg). At this pressure level, it's easier to predict how gases will act.
By using STP conditions, scientists and students can communicate their findings universally, as these conditions are the agreed-upon benchmarks for gas comparisons.
Whenever dealing with gas law calculations, such as determining volumes or pressure changes, remembering the STP values is fundamental to achieving accurate results.
Volume calculation
Calculating the volume of a gas under varying conditions involves understanding how temperature, pressure, and volume interplay, often guided by the Combined Gas Law. This equation is derived from Boyle's, Charles's, and Gay-Lussac's laws, combining them to form:
\[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \]Where:
  • **\( P_1 \)** and **\( P_2 \)** represent initial and final pressures, respectively
  • **\( V_1 \)** and **\( V_2 \)** stand for the initial and final volumes of the gas
  • **\( T_1 \)** and **\( T_2 \)** are the initial and final temperatures in Kelvin
This formula tells you how to relate the conditions of a gas sample before and after it undergoes a change. Solving for the final volume \( V_2 \) is a matter of rearranging the formula and substituting the known values, as was shown in our solution. This allows you to solve for the unknown variable based on how the other factors change.
Temperature conversion to Kelvin
Converting temperatures to Kelvin is a key step in using the Combined Gas Law, as temperatures must be in Kelvin for the calculations to be accurate. Unlike Celsius, Kelvin starts at absolute zero, meaning no negative values. This avoids complications in calculations.
Conversion is simple: just add 273 to the Celsius value.
For example, with an initial temperature of 35掳C:- The conversion is: \[ T(K) = T(掳C) + 273 \]- Thus: \( 35 + 273 = 308 \) KelvinThis conversion ensures that you'll have a temperature value that can be consistently used in thermodynamics and gas law equations without producing negative numbers that could cause errors in calculation.

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Most popular questions from this chapter

A sample of air contains only nitrogen and oxygen gases whose partial pressures are 0.80 atm and 0.20 atm, respectively. Calculate the total pressure and the mole fractions of the gases.

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Explain why a helium weather balloon expands as it rises in the air. Assume that the temperature remains constant

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