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Chapter 14: Problem 87

One mole of a gas enclosed in a vessel is heated at constant pressure \(1 \mathrm{~K}\). Work done by the gas is (a) \(1 \mathrm{~J}\) (b) \(\frac{1}{R} \mathrm{~J}_{1}\) (c) \(R \mathrm{~J}\) (d) None of these

Short Answer

Expert verified
The work done by the gas is option (c): \( R \) Joules.

Step by step solution

01

Understanding the Problem

We are tasked with finding the work done by one mole of gas when it is heated at constant pressure and experiences a temperature change of 1 Kelvin.
02

Reviewing the Formula for Work Done by a Gas

For an ideal gas undergoing an isobaric (constant pressure) process, the work done by the gas can be calculated using the formula: \[ W = P \Delta V \] where \( W \) is the work done, \( P \) is the constant pressure, and \( \Delta V \) is the change in volume.
03

Relating Pressure, Volume, and Temperature for Ideal Gas

Using the Ideal Gas Law, \( PV = nRT \), at constant pressure, the change in volume, \( \Delta V \), when temperature changes by \( \Delta T \) is:\[ \Delta V = \frac{nR \Delta T}{P} \] where \( n = 1 \) mole, \( R \) is the ideal gas constant, and \( \Delta T = 1 \) K.
04

Calculating the Work Done

Substitute \( \Delta V \) into the work formula:\[ W = P \times \Delta V = P \times \left(\frac{nR \Delta T}{P}\right) = nR \Delta T \]Since \( n = 1 \) and \( \Delta T = 1 \) K, the work done simplifies to:\[ W = 1 \times R \times 1 = R \]
05

Comparing with Given Options

The calculated work done, \( R \), matches option (c), which is \( R \) Joules.

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Key Concepts

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

work done by gas
When a gas expands or contracts, it either does work on the surroundings or has work done on it. This concept is essential for understanding energy transfer in systems involving gases. In thermodynamics, the work done by or on a gas is calculated based on the pressure and volume changes of the gas. This is expressed through the formula:
  • The work done, \( W \), can be expressed as: \( W = P \Delta V \), where \( P \) is the pressure and \( \Delta V \) is the change in volume.
  • This formula is applicable specifically for an isobaric process, where the pressure \( P \) remains constant.
In a practical scenario, consider heating a gas at constant pressure. The gas will expand, increasing its volume. This expansion implies the gas has done work on its surroundings. Since the pressure is constant, the more the volume increases, the more work is done.
ideal gas law
The Ideal Gas Law is a fundamental equation in physics that helps describe the behavior of an ideal gas. The equation is written as:
  • \( PV = nRT \)
  • where:\*\( P \) is the pressure,\*\( V \) is the volume,\*\( n \) is the number of moles of the gas,\*\( R \) is the ideal gas constant, and\*\( T \) is the temperature in Kelvin.
This equation combines three different gas laws: Boyle's Law, Charles's Law, and Avogadro's Law. It provides a powerful tool when you need to calculate one of the state variables of the gas (pressure, volume, or temperature) if the others are known.
In our context of continuous heating at constant pressure, as the temperature changes, the volume changes in such a way that the Ideal Gas Law is maintained consistent. This allows us to relate the work done to temperature change.
isobaric process
An isobaric process is a thermodynamic process during which the pressure stays constant. This is a common type of process in thermodynamics and can be visualized as a horizontal line on a pressure-volume (P-V) diagram.
  • In this specific exercise, since the process is isobaric, it means the pressure does not change when the gas is heated.
  • Any area under the P-V curve on such diagrams represents the work done by the gas during its expansion or contraction.
The advantage of dealing with an isobaric process is that calculations become relatively simple. For instance, if we know the temperature change and the number of moles, using the Ideal Gas Law and the work formula, we can efficiently calculate the work done by or on the gas. This characteristic of simplicity derives from the constant pressure condition, which narrows down the variables involved.

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Most popular questions from this chapter

In an adiabatic change, the pressure and temperature of a monoatomic gas are related as \(p \propto T^{-c}\) where \(c\) equals (a) \(\frac{2}{5}\) (b) \(\frac{5}{2}\) (c) \(\frac{3}{5}\) (d) \(\frac{5}{3}\)

Three copper blocks of masses \(M_{1}, M_{2}\) and \(M_{3} \mathrm{~kg}\) respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at \(T_{1}, T_{2}, T_{3}\left(T_{1}>T_{2}>T_{3}\right) .\) Assuming there is no heat loss to the surroundings, the equilibrium temprature \(T\) is ( \(s\) is specific heat of copper) [NCERT Exemplar] (a) \(T=\frac{T_{1}+T_{2}+T_{3}}{3}\) (b) \(T=\frac{M_{1} T_{1}+M_{2} T_{2}+M_{3} T_{3}}{M_{1}+M_{2}+M_{3}}\) (c) \(T=\frac{M_{1} T_{1}+M_{2} T_{2}+M_{3} T_{3}}{3\left(M_{1}+M_{2}+M_{3}\right)}\) (d) \(T=\frac{M_{1} T_{1} s+M_{2} T_{2} s+M_{3} T_{3}}{M_{1}+M_{2}+M_{3}}\)

For a gas, the difference between the two principal specific heats is \(4150 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}\). What is the specific heat of the gas at constant volume if, the ratio of specific heat is \(1.4 ?\) (a) \(5186 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}\) (b) \(10375 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}\) (c) \(1660 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}\) (d) \(8475 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}\)

How many times a diatomic gas should be expanded adiabatically so as to reduce the root mean square velocity to half (a) 64 (b) 32 (c) 16 (d) 8

A Carnot engine used first ideal monoatomic gas and then an ideal diatomic gas, if the source and sink temperatures are \(411^{\circ} \mathrm{C}\) and \(69^{\circ} \mathrm{C}\), respectively and the engine extracts \(1000 \mathrm{~J}\) of heat from the source in each cycle, then (a) area enclosed by the \(p-V\) diagram is \(10 \mathrm{~J}\) (b) heat energy rejected by engine is 1 st case is \(600 \mathrm{~J}\) while that in 2 nd case in \(113 \mathrm{~J}\) (c) area enclosed by the \(p-V\) diagram is \(500 \mathrm{~J}\) (d) efficiencies of the engine in both the cases are in the ratio \(21: 25\)
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