/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Molecular hydrogen gas having a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Molecular hydrogen gas having a mass of \(6.44 \mathrm{~g}\) at \(26.0^{\circ} \mathrm{C}\) is heated until its volume doubles while it is held at a constant pressure. How much work was done by the gas? [Hint: Take it to be an ideal gas.]

Short Answer

Expert verified
The work done by the gas is approximately 7940.5 J.

Step by step solution

01

Understand the Problem

The problem asks us to find the work done by a gas when its volume is doubled at constant pressure, given the mass and initial temperature of molecular hydrogen (assumed to be an ideal gas).
02

Identify Relevant Formula and Variables

For an ideal gas at constant pressure, the work done by the gas when its volume changes from \(V_1\) to \(V_2\) is given by \(W = P \Delta V\), where \(\Delta V = V_2 - V_1\). To find \(V_1\) and \(V_2\), use the ideal gas law, \(PV = nRT\).
03

Calculate the Number of Moles (n)

Molar mass of hydrogen (\(H_2\)) is approximately \(2.02 \mathrm{~g/mol}\). Number of moles, \(n = \frac{6.44 \mathrm{~g}}{2.02 \mathrm{~g/mol}} \approx 3.19 \mathrm{~mol}\).
04

Calculate Initial Volume (V1)

Using the ideal gas law, \(PV_1 = nRT_1\). Convert initial temperature to Kelvin: \(T_1 = 26.0^{\circ}C + 273.15 = 299.15 \text{ K}\). Rearrange to find \(V_1 = \frac{nRT_1}{P}\).
05

Calculate Final Volume (V2)

Since the volume doubles, \(V_2 = 2V_1\).
06

Find Work Done by the Gas

The work done \(W\) is \(P\Delta V = P(V_2 - V_1) = P(2V_1 - V_1) = PV_1\). Substitute \(PV_1 = nRT_1\) from the ideal gas law, and \[ W = nRT_1 \]. Use \(R = 8.314 \mathrm{~J/(mol \cdot K)}\) to calculate \[ W = 3.19 \times 8.314 \times 299.15 \approx 7940.5 \mathrm{~J} \].
07

Conclusion

Thus, the work done by the gas when its volume is doubled at constant pressure is approximately \(7940.5 \mathrm{~J}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 thermodynamics that describes the behavior of an ideal gas. This law combines several gas laws to form a single equation: \[ PV = nRT \]where:
  • \(P\) is the pressure of the gas, typically measured in pascals (Pa).
  • \(V\) is the volume of the gas in cubic meters (m\(^3\)).
  • \(n\) is the number of moles of gas.
  • \(R\) is the ideal gas constant, approximately \(8.314 \text{ J/(mol} \cdot \text{K)}\).
  • \(T\) is the temperature of the gas in Kelvin (K).
This equation relates these variables, making it possible to calculate one if the others are known. For example, in our exercise, it helps us derive how the volume changes with temperature when pressure is held constant.
Work done by gas
Work done by a gas during expansion or compression is a crucial concept in thermodynamics. Whenever a gas changes its volume under constant pressure, it performs work. The formula to calculate work done by the gas is:\[ W = P \, \Delta V \]Here, \( \Delta V \) is the change in volume, which can be expressed as \( V_2 - V_1 \).
  • When the volume increases, the gas does positive work on the surroundings.
  • Conversely, when the volume decreases, work is done on the gas.
  • In our specific scenario, given that the volume doubles at constant pressure, the work done can be efficiently determined by calculating \( nRT_1 \) as demonstrated. This simplification arises from substituting the ideal gas conditions into the work formula.
    Moles calculation
    The calculation of moles is an essential step in solving problems involving the ideal gas law. Moles (represented as \( n \)) indicate the amount of substance within the gas. Using the molar mass of a substance allows us to convert between mass and moles:\[ n = \frac{\text{mass}}{\text{molar mass}} \]In our problem:
    • The molar mass of hydrogen gas (\( H_2 \)) is approximately \(2.02 \text{ g/mol}\).
    • The given mass of hydrogen is \(6.44 \text{ g}\).
    Thus, the number of moles is calculated as:\[ n = \frac{6.44 \text{ g}}{2.02 \text{ g/mol}} \approx 3.19 \text{ mol} \]This value is crucial for pinpointing the volume and pressure in further calculations using the ideal gas law.
    Constant pressure processes
    Constant pressure processes, also known as isobaric processes, play a vital role in thermodynamic studies. These processes occur when the pressure remains unchanged while the other factors—such as volume and temperature—may vary. For gases, the relationship between temperature and volume at constant pressure is governed by Charles's Law:\[ V \propto T \]or, in an equation form: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]This principle aids in understanding how gas expands or contracts under temperature alteration at constant pressure.In the case we discussed, as the temperature of the gas increased, its volume doubled.
    • This doubling of volume meant the work done by the gas could be calculated using the ideal gas law's conditions.
    Knowing these details allows us to appreciate how the gas performs work when subjected to a rise in temperature capable of maintaining constant pressure through its expansion.

    One App. One Place for Learning.

    All the tools & learning materials you need for study success - in one app.

    Get started for free

    Most popular questions from this chapter

    To determine the specific heat of an oil, an electrical heating coil is placed in a calorimeter with 380 g of the oil at \(10^{\circ} \mathrm{C}\). The coil consumes energy (and gives off heat) at the rate of \(84 \mathrm{~W}\). After \(3.0 \mathrm{~min}\), the oil temperature is \(40^{\circ} \mathrm{C}\). If the water equivalent of the calorimeter and coil is \(20 \mathrm{~g}\), what is the specific heat of the oil?

    For nitrogen gas, \(c_{v}=740 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Assuming it to behave like an ideal gas, find its specific heat at constant pressure. (The molecular mass of nitrogen gas is \(28.0 \mathrm{~kg} / \mathrm{kmol} .\) ) Method 1 $$c_{p}=c_{v}+\frac{R}{M}=\frac{740 \mathrm{~J}}{\mathrm{~kg} \cdot \mathrm{K}}+\frac{8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K}}{28.0 \mathrm{~kg} / \mathrm{kmol}}=1.04 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$$ Method 2 Since \(\mathrm{N}_{2}\) is a diatomic gas, and since \(\gamma=c_{p} / c_{v}\) for such a gas, $$ c_{p}=1.40 c_{v}=1.40(740 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})=1.04 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} $$

    A gas at a pressure of \(2.10 \times 10^{5}\) Pa occupies \(4.98 \times 10^{-3} \mathrm{~m}^{3}\) in a chamber that can change its volume. The gas is at an initial temperature of \(290 \mathrm{~K}\) when it is heated, so that it expands isobarically, thereupon doing \(200 \mathrm{~J}\) of work. Determine the new volume and the final temperature of the gas. [Hint: Use Eq. (20.1) and the Ideal Gas Law applied before and after the expansion.]

    A spring \((k=500 \mathrm{~N} / \mathrm{m})\) supports a 400 -g mass, which is immersed in 900 g of water. The specific heat of the mass is \(450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The spring is now stretched \(15 \mathrm{~cm}\), and after thermal equilibrium is reached, the mass is released so it vibrates up and down. By how much has the temperature of the water changed when the vibration has stopped? The energy stored in the spring is dissipated by the effects of friction and goes to heat the water and mass. The energy stored in the stretched spring was $$\mathrm{PE}_{e}=\frac{1}{2} k x^{2}=\frac{1}{2}(500 \mathrm{~N} / \mathrm{m})(0.15 \mathrm{~m})^{2}=5.625 \mathrm{~J}$$ This energy appears as thermal energy that flows into the water and the mass. Using \(\Delta Q=\mathrm{cm} \Delta T\), $$5.625 \mathrm{~J}=(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(0.900 \mathrm{~kg}) \Delta T+(450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(0.40 \mathrm{~kg}) \Delta 7$$ which leads to $$ \Delta T=\frac{5.625 \mathrm{~J}}{3950 \mathrm{~J} / \mathrm{K}}=0.0014 \mathrm{~K} $$

    In a certain process, \(8.00\) kcal of heat is furnished to the system while the system does \(6.00 \mathrm{~kJ}\) of work. By how much does the internal energy of the system change during the process? Here \(8.00\) kcal is heat-in and \(6.00 \mathrm{~kJ}\) is work-out, both of which are positive. Consequently, \(\Delta Q=(8000\) cal \()(4.184 \mathrm{~J} / \mathrm{cal})=33.5 \mathrm{~kJ}\) and \(\Delta W=6.00 \mathrm{~kJ}\) Therefore, from the First Law \(\Delta Q=\Delta U+\Delta W\), $$\Delta U=\Delta Q-\Delta W=33.5 \mathrm{~kJ}-6.00 \mathrm{~kJ}=27.5 \mathrm{~kJ}$$

    See all solutions

    Recommended explanations on Physics Textbooks

    View all explanations

    What do you think about this solution?

    We value your feedback to improve our textbook solutions.

    Study anywhere. Anytime. Across all devices.