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Physics Chapter 8 : Topic 8
First Law of Thermodynamics
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The first law of thermodynamics is a statement of:
Reasoning: The first law states that heat supplied to a system is used to increase its internal energy and perform work, showing conservation of energy.
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According to the first law of thermodynamics, the heat supplied to a system is equal to:
Reasoning: The first law is expressed as $\mathbf{Q = \mathrm{\Delta}U + W}$, where $\mathbf{Q}$ is heat supplied, $\mathbf{\mathrm{\Delta}U}$ is increase in internal energy, and $\mathbf{W}$ is work done on the system.
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In the equation $\mathbf{Q = \mathrm{\Delta}U + W}$, the symbol $\mathbf{Q}$ represents:
Reasoning: In this sign convention, $\mathbf{Q}$ represents energy transferred to the system by heating.
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A gas absorbs $\mathbf{500\ J}$ of heat and performs $\mathbf{200\ J}$ of work on the surroundings. The increase in internal energy is:
Reasoning: Work done by gas means energy leaves the system.Therefore, $\mathbf{\mathrm{\Delta}U = Q - W = 500 - 200 = 300\ J}$
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If heat is supplied to a gas and no work is done, then:
Reasoning: When $\mathbf{W = 0}$, the first law becomes $\mathbf{Q = \mathrm{\Delta}U + 0}$, so all heat increases internal energy.
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A system receives $\mathbf{800\ J}$ of heat and $\mathbf{300\ J}$ of work is done on it. The increase in internal energy is:
Reasoning: Using $\mathbf{Q = \mathrm{\Delta}U + W}$, $\mathbf{\mathrm{\Delta}U = Q - W}$. Since work is done on the system, it is included as energy input: $\mathbf{\mathrm{\Delta}U = 800 + 300 = 1100\ J}$.
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Which situation causes an increase in internal energy of a system?
Reasoning: Both heating and work done on the system add energy, increasing internal energy.
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If $\mathbf{Q = 0}$ for an isolated system, then:
Reasoning: For an adiabatic process, $\mathbf{Q = 0}$. From $\mathbf{Q = \mathrm{\Delta}U + W}$, $\mathbf{0 = \mathrm{\Delta}U + W}$, so $\mathbf{\mathrm{\Delta}U = - \ W}$.
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In thermodynamics, internal energy mainly depends on:
Reasoning: Internal energy is the total microscopic kinetic and potential energy of molecules.
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A gas expands and does $\mathbf{400\ J}$ of work while receiving $\mathbf{600\ J}$ heat. The change in internal energy is:
Reasoning: Heat supplied is used for work and internal energy:$\mathbf{\mathrm{\Delta}U = Q - W = 600 - 400 = 200\ J}$
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When work is done on a system, according to the given convention:
Reasoning: Work done on the system transfers mechanical energy into the system.
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A system loses $\mathbf{300\ J}$ of heat and $\mathbf{100\ J}$ work is done on it. The change in internal energy is:
Reasoning: Heat loss means $\mathbf{Q = - \ 300\ J}$. Work input means $\mathbf{W = 100\ J}$.$\mathbf{Q = \mathrm{\Delta}U + W}$ $\mathbf{\Longrightarrow}$ $\mathbf{- \ 300 = \mathrm{\Delta}U + 100}$ $\mathbf{\Longrightarrow}$ $\mathbf{\mathrm{\Delta}U = - \ 400\ J}$
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During an isothermal process of an ideal gas:
Reasoning: For an ideal gas, internal energy depends only on temperature.Constant temperature means $\mathbf{\mathrm{\Delta}U = 0}$.
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For an isothermal expansion of an ideal gas, the heat supplied is:
Reasoning: Since $\mathbf{\mathrm{\Delta}U = 0}$, the first law gives $\mathbf{Q = W}$.
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The unit of $\mathbf{Q\ }$, $\mathbf{W}$ and $\mathbf{\mathrm{\Delta}U}$ in SI system is:
Reasoning: Heat, work, and internal energy are all forms of energy measured in joules.
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A system absorbs $\mathbf{1000\ J}$ heat and its internal energy increases by $\mathbf{600\ J}$. The work done is:
Reasoning: Using, $\mathbf{Q = \mathrm{\Delta}U + W}$ $\mathbf{\Longrightarrow}$ $\mathbf{1000 = 600 + W}$ $\mathbf{\Longrightarrow}$ $\mathbf{W = 400\ J}$
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A system receives no heat but its internal energy increases. This means:
Reasoning: If $\mathbf{Q = 0}$ and $\mathbf{\mathrm{\Delta}U}$ increases, energy must come from work done on the system.
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A gas is compressed without heat exchange. The work done on gas will:
Reasoning: In adiabatic compression, work input increases molecular energy, raising internal energy.
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Which process has $\mathbf{Q = 0}$?
Reasoning: An adiabatic process occurs without heat exchange between system and surroundings.
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In a cyclic process, the final and initial states are the same. Therefore:
Reasoning: Internal energy is a state function. Returning to the original state means no net change in internal energy.
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If a system completes a cycle and absorbs $\mathbf{500\ J}$ heat, the net work done by the system is:
Reasoning: For a cycle, $\mathbf{\mathrm{\Delta}U = 0}$. Therefore, $\mathbf{Q = W}$.
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Which quantity in the first law represents energy stored inside the system?
Reasoning: $\mathbf{\mathrm{\Delta}U}$ represents the change in internal energy stored within the system.
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A gas receives $\mathbf{200\ J}$ heat and $\mathbf{50\ J}$ work is done by the gas. The remaining energy:
Reasoning: $\mathbf{\mathrm{\Delta}U = Q - W = 200 - 50 = 150\ J}$
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Which statement is correct about heat and work?
Reasoning: Heat and work are energy transfer mechanisms, not stored properties.
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The first law of thermodynamics rejects the possibility of:
Reasoning: The law states energy cannot be created or destroyed; it can only be transferred or transformed.
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A gas is heated while kept at constant volume. The work done is:
Reasoning: At constant volume, there is no displacement, so no work is done.
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If all supplied heat is converted into work, then:
Reasoning: If $\mathbf{Q = W}$, then from $\mathbf{Q = \mathrm{\Delta}U + W}$, $\mathbf{\mathrm{\Delta}U = 0}$.
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If the internal energy of a system decreases while it absorbs heat, then:
Reasoning: If $\mathbf{\mathrm{\Delta}U}$ is negative while $\mathbf{Q}$ is positive, the system must lose more energy through work than it gains by heating.
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The first law of thermodynamics connects:
Reasoning: The first law establishes the relationship between heat transfer, work, and change in internal energy.
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