/*! 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 79 ssM A sample of gas expands from... [FREE SOLUTION] | 91Ó°ÊÓ

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ssM A sample of gas expands from an initial pressure and volume of \(10 \mathrm{~Pa}\) and \(1.0 \mathrm{~m}^{3}\) to a final volume of \(2.0 \mathrm{~m}^{3}\). During the expansion, the pressure and volume are related by the equation \(p=a V^{2}\), where \(a=10 \mathrm{~N} / \mathrm{m}^{8} .\) Determine the work done by the gas during this expansion.

Short Answer

Expert verified
The work done by the gas is 23.33 Joules.

Step by step solution

01

Understand the Relationship

The pressure-volume relationship is given by the equation \( p = a V^2 \), where \( a = 10 \, \text{N/m}^8 \). This means pressure \( p \) changes with the square of volume \( V \). For any specific volume \( V \), pressure can be calculated using this formula.
02

Set Up the Integral for Work Done

The work done by a gas during expansion can be found by calculating the integral of pressure with respect to volume: \( W = \int_{V_1}^{V_2} p \, dV \). Substitute \( p = a V^2 \) to get \( W = \int_{V_1}^{V_2} a V^2 \, dV \).
03

Calculate the Integral

Substitute the known value of \( a \) into the integral: \( W = \int_{1}^{2} 10 V^2 \, dV \). Solve the integral: \[W = 10 \int_{1}^{2} V^2 \, dV = 10 \left[ \frac{V^3}{3} \right]_{1}^{2}.\]
04

Evaluate the Definite Integral

Substitute the limits of integration into the solved integral:\[W = 10 \left( \frac{2^3}{3} - \frac{1^3}{3} \right) = 10 \left( \frac{8}{3} - \frac{1}{3} \right) = 10 \left( \frac{7}{3} \right).\]This simplifies to: \( W = \frac{70}{3} \).
05

Calculate the Final Work Done

Calculate \( \frac{70}{3} \), which equals \( 23.33 \). Thus, the work done by the gas during expansion is \( 23.33 \, \text{Joules} \).

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Key Concepts

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

Pressure-volume relationship
The relationship between pressure and volume is crucial in understanding how gases behave during processes like expansion. In our exercise, the equation \( p = a V^2 \) describes the pressure-volume relationship. Here, pressure \( p \) is dependent on the volume \( V \) squared, and the constant \( a \) determines the proportionality. This means that as the volume increases, the pressure increases at a rate proportional to the square of the volume. This quadratic relationship is typical in scenarios where pressure is not constant, such as during specific types of thermodynamic processes.

Understanding this relationship helps us make predictions about how the gas will behave as it expands or contracts. It is important for comprehending physical phenomena like breathing, engine operations, and even weather patterns. In these contexts, knowing how to relate volume changes to pressure changes allows scientists and engineers to effectively model and control systems.
Integral calculus in physics
Integral calculus is a fundamental tool in physics for calculating quantities like work, area, and accumulated change over a range. In our exercise, we used integral calculus to determine the work done by the gas. The formula \( W = \int_{V_1}^{V_2} p \, dV \) is essential to finding the work done during an expansion where pressure varies with volume.

By substitution, since \( p = a V^2 \), we integrate \( W = \int_{1}^{2} a V^2 \, dV \) with \( a \) as a constant factor, leading us to \( W = 10 \int_{1}^{2} V^2 \, dV \). Calculating this integral involves finding the antiderivative of \( V^2 \), which is \( \frac{V^3}{3} \).

Evaluating the definite integral from the initial volume to the final volume, helps to determine the net amount of work done during expansion. Hence, integral calculus serves as a bridge between theoretical concepts like pressure-volume relations and practical calculations necessary for solving real-world problems.
Thermodynamics expansion
Thermodynamic expansion refers to the increase in volume of a gas or substance when temperature or pressure conditions change. In this exercise, we deal with the expansion of a gas from a volume of \(1.0 \, \text{m}^3\) to \(2.0 \, \text{m}^3\).

In thermodynamics, expansion can be isothermal, isobaric, adiabatic, or occur under varying conditions like ours. Here, because pressure changes according to \( p = a V^2 \), the expansion is not one that fits into standard categories like isothermal or adiabatic, but it requires integral calculus to understand how work is done.

Studying such expansions gives insight into how systems like engines work. It helps to determine the amount of energy necessary for processes like compression and expansion cycles. Thus, understanding thermodynamic expansions is key to improving energy efficiency and designing sustainable systems.

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Most popular questions from this chapter

The specific heat of a substance varies with temperature according to the function \(c=0.20+0.14 T+0.023 T^{2}\), with \(T\) in \({ }^{\circ} \mathrm{C}\) and \(c\) in \(\mathrm{cal} / \mathrm{g} \cdot \mathrm{K}\). Find the energy required to raise the temperature of \(2.0\) g of this substance from \(5.0^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\).

A 150 g copper bowl contains 220 g of water, both at \(20.0^{\circ} \mathrm{C}\). A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with \(5.00 \mathrm{~g}\) being converted to steam. The final temperature of the system is \(100^{\circ} \mathrm{C}\). Neglect energy transfers with the environment. (a) How much energy (in calories) is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder?

\(.\) Nonmetric version: (a) How long does a \(2.0 \times 10^{5} \mathrm{Btu} / \mathrm{h}\) water heater take to raise the temperature of 40 gal of water from \(70^{\circ} \mathrm{F}\) to \(100^{\circ} \mathrm{F}\) ? Metric version: (b) How long does a \(59 \mathrm{~kW}\) water heater take to raise the temperature of \(150 \mathrm{~L}\) of water from \(21^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C} ?\)

(a) Two 50 g ice cubes are dropped into \(200 \mathrm{~g}\) of water in a thermally insulated container. If the water is initially at \(25^{\circ} \mathrm{C}\), and the ice comes directly from a freezer at \(-15^{\circ} \mathrm{C}\), what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used?

A certain diet doctor encourages people to diet by drinking ice water. His theory is that the body must burn off enough fat to raise the temperature of the water from \(0.00^{\circ} \mathrm{C}\) to the body temperature of \(37.0^{\circ} \mathrm{C}\). How many liters of ice water would have to be consumed to burn off \(454 \mathrm{~g}\) (about \(1 \mathrm{lb}\) ) of fat, assuming that burning this much fat requires 3500 Cal be transferred to the ice water? Why is it not advisable to follow this diet? (One liter \(=10^{3} \mathrm{~cm}^{3}\). The density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3} .\) )

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