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A sample of an ideal gas is taken through the cyclic process “abcd ” as shown in figure. It absorbed 50Jof heat during part ab, no heat during bc and reject 70J of heat during ca, 40J of work is done on the gas during part bc .(a) Find the internal energy of the gas at b and c if it is 1500 J at a .
? Related: Zeroth & First Law of Thermodynamics
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A sample of an ideal gas is taken through the cyclic process “abcd ” a...
Zeroth Law of Thermodynamics
The zeroth law of thermodynamics states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law establishes the concept of temperature and allows us to define a temperature scale.
First Law of Thermodynamics
The first law of thermodynamics, also known as the law of energy conservation, states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. Mathematically, it is represented as ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
Analysis of the Cyclic Process
- Internal energy at point a: Ua = 1500 J
- Heat absorbed during ab: Qab = 50 J
- Heat rejected during ca: Qca = 70 J
- Work done during bc: Wbc = -40 J (negative because work is done on the gas)
- Internal energy at point b: Ub = Ua + Qab + Wbc = 1500 J + 50 J - 40 J = 1510 J
- Internal energy at point c: Uc = Ub + Qbc = 1510 J (no heat transfer during bc)
Therefore, the internal energy of the gas at point b is 1510 J and at point c is also 1510 J. This analysis follows the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system minus the work done by the system.
The zeroth law of thermodynamics states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law establishes the concept of temperature and allows us to define a temperature scale.
First Law of Thermodynamics
The first law of thermodynamics, also known as the law of energy conservation, states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. Mathematically, it is represented as ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
Analysis of the Cyclic Process
- Internal energy at point a: Ua = 1500 J
- Heat absorbed during ab: Qab = 50 J
- Heat rejected during ca: Qca = 70 J
- Work done during bc: Wbc = -40 J (negative because work is done on the gas)
- Internal energy at point b: Ub = Ua + Qab + Wbc = 1500 J + 50 J - 40 J = 1510 J
- Internal energy at point c: Uc = Ub + Qbc = 1510 J (no heat transfer during bc)
Therefore, the internal energy of the gas at point b is 1510 J and at point c is also 1510 J. This analysis follows the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system minus the work done by the system.
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A sample of an ideal gas is taken through the cyclic process “abcd ” as shown in figure. It absorbed 50Jof heat during part ab, no heat during bc and reject 70J of heat during ca, 40J of work is done on the gas during part bc .(a) Find the internal energy of the gas at b and c if it is 1500 J at a . Related: Zeroth & First Law of Thermodynamics?
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A sample of an ideal gas is taken through the cyclic process “abcd ” as shown in figure. It absorbed 50Jof heat during part ab, no heat during bc and reject 70J of heat during ca, 40J of work is done on the gas during part bc .(a) Find the internal energy of the gas at b and c if it is 1500 J at a . Related: Zeroth & First Law of Thermodynamics? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A sample of an ideal gas is taken through the cyclic process “abcd ” as shown in figure. It absorbed 50Jof heat during part ab, no heat during bc and reject 70J of heat during ca, 40J of work is done on the gas during part bc .(a) Find the internal energy of the gas at b and c if it is 1500 J at a . Related: Zeroth & First Law of Thermodynamics? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample of an ideal gas is taken through the cyclic process “abcd ” as shown in figure. It absorbed 50Jof heat during part ab, no heat during bc and reject 70J of heat during ca, 40J of work is done on the gas during part bc .(a) Find the internal energy of the gas at b and c if it is 1500 J at a . Related: Zeroth & First Law of Thermodynamics?.
A sample of an ideal gas is taken through the cyclic process “abcd ” as shown in figure. It absorbed 50Jof heat during part ab, no heat during bc and reject 70J of heat during ca, 40J of work is done on the gas during part bc .(a) Find the internal energy of the gas at b and c if it is 1500 J at a . Related: Zeroth & First Law of Thermodynamics? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A sample of an ideal gas is taken through the cyclic process “abcd ” as shown in figure. It absorbed 50Jof heat during part ab, no heat during bc and reject 70J of heat during ca, 40J of work is done on the gas during part bc .(a) Find the internal energy of the gas at b and c if it is 1500 J at a . Related: Zeroth & First Law of Thermodynamics? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample of an ideal gas is taken through the cyclic process “abcd ” as shown in figure. It absorbed 50Jof heat during part ab, no heat during bc and reject 70J of heat during ca, 40J of work is done on the gas during part bc .(a) Find the internal energy of the gas at b and c if it is 1500 J at a . Related: Zeroth & First Law of Thermodynamics?.
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