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Chapter 15: 15-5Q (page 412)

An ideal monatomic gas expands slowly to twice its volume (1) isothermally; (2) adiabatically; (3) isobarically. Plot each on a PV diagram. In which process \(\Delta U\) is the greatest, and in which is \(\Delta U\) the least? In which is W the greatest and the least? In which is Q the greatest and the least?

Short Answer

Expert verified

The change in internal energy is the greatest is for the isobaric process and the least for the adiabatic process.

The work done is the greatest in the isobaric process and the least for the adiabatic process.

Q is the greatest for the isobaric process and the least for the adiabatic process.

Step by step solution

01

Concept

By the first law of thermodynamics, \(\Delta U = Q - W\).

The work is the area under the curve in the P-V diagram.

02

Explanation for \(\Delta U\)

For the isothermal process, the temperature change and internal energy change are zero.

The change in the internal energy is the greatest for the isobaric process and the least for the adiabatic process.

03

Explanation for \(W\)

The work is the area under the curve in the P-V diagram. Here, the area is more for the isobaric process and less for the adiabatic process.

Hence, in the isobaric process, the work done is the greatest and the least for the adiabatic process.

04

Explanation for \(Q\)

For the adiabatic process, the change in the heat energy is zero.

Heat energy equals the sum of the work done and the change in the system's internal energy.

The sum is maximum for the isobaric process.

Hence, Q is the greatest for the isobaric process and the least for the adiabatic process.

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(II) When\({\bf{5}}{\bf{.80 \times 1}}{{\bf{0}}{\bf{5}}}\;{\bf{J}}\)of heat is added to a gas enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure, the volume is observed to increase from\({\bf{1}}{\bf{.9}}\;{{\bf{m}}{\bf{3}}}\)to\({\bf{4}}{\bf{.1}}\;{{\bf{m}}{\bf{3}}}\). Calculate

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(c) Graph this process on a PV diagram.
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