A gas in a cylinder was placed in a heater and gained \(7000 \mathrm{~kJ}\) of heat. If the cylinder increased in volume from \(700 \mathrm{~mL}\) to \(1450 \mathrm{~mL}\) against an atmospheric pressure of 750 Torr during this process, what is the change in internal energy of the gas in the cylinder?

Short Answer

Expert verified
The change in internal energy is 6992.25 kJ

Step by step solution

01

Convert Volume Change to Liters

First, convert the change in volume from milliliters to liters. To do this, use the fact that 1 liter equals 1000 milliliters. The change in volume (ΔV) in liters is calculated as the final volume minus the initial volume both converted to liters.
02

Convert Pressure to the SI Unit

Convert the pressure from Torr to the SI unit of pascals (Pa). Since there are 133.322 Pa in 1 Torr, you multiply the given pressure by this conversion factor to find the pressure in pascals.
03

Calculate the Work Done

The work done (W) on the gas is calculated using the equation W = -PΔV, where P is the pressure in pascals and ΔV is the change in volume in liters (but must be converted to cubic meters for the calculation, by multiplying by 1e-3). The negative sign indicates that work is done by the system when it expands.
04

Convert the Work from Joules to Kilojoules

After calculating the work done in joules, convert it to kilojoules by dividing by 1000, as 1 kJ equals 1000 J.
05

Calculate the Change in Internal Energy

Use the first law of thermodynamics, which states that the change in internal energy (ΔU) equals the heat added to the system (Q) minus the work done by the system (W). Substitute the given heat in kilojoules and the calculated work (converted to kilojoules) into the equation ΔU = Q - W to find the change in internal energy.

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 Vaia!

Key Concepts

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

First Law of Thermodynamics
The first law of thermodynamics, often referred to as the law of energy conservation, is a fundamental principle that describes the relationship between the heat added to a system and the work done by the system. It asserts that the change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the work done by the system on its surroundings.

This law can be mathematically represented as \( \Delta U = Q - W \), where \( \Delta U \) stands for the change in internal energy, \( Q \) for the heat added to the system, and \( W \) for the work done by the system. It is a pivotal concept in physics and chemistry that helps us understand how different forms of energy are interconvertible and the energy balance within a system.
Work Done by Gas
When we talk about 'work done by gas' in thermodynamics, we refer to the energy transferred when a gas expands or contracts within a container. To calculate the work done by the gas during expansion or compression, the following equation is often used: \( W = -P \Delta V \), where \( W \) is the work done by the gas, \( P \) is the pressure, and \( \Delta V \) is the change in volume of the gas. It is important to note that work is considered positive when done on the system and negative when done by the system.

This concept is significant because it connects mechanical work with thermal processes. Understanding how to calculate work is critical for predicting how systems will behave when energy is added or removed. In the context of an expanding gas, as in our exercise, work done by the system on its surroundings occurs as the gas pushes against an external pressure to increase its volume.
Heat Transfer
Heat transfer is a process by which energy, due to a temperature difference, is exchanged either within a system or between different systems. There are three modes of heat transfer: conduction, convection, and radiation. In the context of our exercise, we are concerned with the amount of heat acquired by the gas when placed in a heater.

The measure of this heat, usually in joules or kilojoules, can directly influence the system's temperature and phase. It's essential when calculating the change in internal energy of a system to know precisely how much heat is added or removed. In the given problem, the gas gains \( 7000 \mathrm{~kJ} \) of heat from the heater, which is a significant factor in determining the resulting change in internal energy.
Gas Expansion
Gas expansion occurs when a gas's volume increases. The gas molecules move faster and spread out, pushing against the confines of their container. This behavior is inherently related to temperature and pressure. When a gas heats up, the increased kinetic energy of the particles causes them to push outwards, leading to expansion.

In our exercise, the expansion is mechanically resisted by the atmospheric pressure, evidenced by the opposing force as the gas does work against this pressure. The difference in volume before and after the heating indicates the degree of expansion. Mathematically, the change in volume \( (\Delta V) \) is a crucial variable both in calculating work done by the gas and understanding the physical changes occurring within the system. The expansion of a gas is not only a fundamental concept in thermodynamics but also a critical element of many engineering and scientific applications, such as in engines and refrigeration systems.

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

Determine the reaction enthalpy for the hydrogenation of ethyne to ethane, \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+\) \(2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\), from the following data: enthalpy of combustion of ethyne, \(-1300 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\); enthalpy of combustion of ethane, \(-1560 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\); enthalpy of combustion of hydrogen, \(-286 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\).

Calculate the enthalpy of the reaction \(\mathrm{P}_{4}(\mathrm{~s})+10 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{PCl}_{5}(\mathrm{~s})\) from the reactions $$ \begin{gathered} \mathrm{P}_{4}(\mathrm{~s})+6 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{3}(\mathrm{I}) \\ \Delta H^{\circ}=-1278.8 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{5}(\mathrm{~s}) \\ \Delta H^{m}=-124 \mathrm{~kJ} \end{gathered} $$

A calorimeter was calibrated with an electric heater, which was used to supply \(22.5 \mathrm{~kJ}\) of energy to the calorimeter; the heat increased the temperature of the calorimeter and its water bath from \(22.45^{\circ} \mathrm{C}\) to \(23.97^{\circ} \mathrm{C}\). What is the heat capacity of the calorimeter?

Near room temperature, the specific heat capacity of benzene is \(1.05 \mathrm{~J} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1} \mathrm{~g}^{-1}\). Calculate the heat nceded to raise the temperature of \(50.0 \mathrm{~g}\) of benzene from \(25.3^{\circ} \mathrm{C}\) to \(37.2^{\circ} \mathrm{C}\). (b) A \(1.0-\mathrm{kg}\) block of aluminum is supplied with \(490 \mathrm{~kJ}\) of heat. What is the temperature change of the aluminum? The specific heat capacity of aluminum is \(0.90 \mathrm{~J} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1} \cdot \mathrm{g}^{-1}\).

A calorimeter has a measured heat capacity of \(6.27 \mathrm{~kJ} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1}\). The combustion of \(1.84 \mathrm{~g}\) of magnesium led to a temperature change from \(21.30^{\circ} \mathrm{C}\) to \(28.56^{\circ} \mathrm{C}\). Calculare the enthalpy change of the reaction \(2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{MgO}(\mathrm{s})\).

See all solutions

Recommended explanations on Chemistry 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.

Sign-up for free