(D) According to the first law of thermodynamics,the heat supplied to a system $(\Delta Q)$ is equal to the sum of the change in internal energy $(\De

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$\Delta U = -\Delta W$ in an adiabatic process

(D) According to the first law of thermodynamics,the heat supplied to a system $(\Delta Q)$ is equal to the sum of the change in internal energy $(\Delta U)$ and the work done by the system $(\Delta W)$: $\Delta Q = \Delta U + \Delta W$ In an adiabatic process,there is no exchange of heat between the system and the surroundings,so $\Delta Q = 0$. Substituting this into the first law equation: $0 = \Delta U + \Delta W$ Therefore,$\Delta U = -\Delta W$.

Class 11 Physics Thermodynamics Adiabatic Process

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In a thermodynamic process,if $\Delta W$ and $\Delta U$ are the work done and the change in the internal energy of a system respectively,then:

AP EAMCET 2023 All AP EAMCET

A
$\Delta U = \Delta W$ in an adiabatic process
B
$\Delta U = \Delta W$ in an isothermal process
C
$\Delta U = -\Delta W$ in an isothermal process
D
$\Delta U = -\Delta W$ in an adiabatic process

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$\Delta U + \Delta W = 0$ is valid for

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$A$ fixed amount of a gas undergoes a thermodynamic process as shown in the figure,such that the heat interaction along path $B \to C \to A$ is equal to the work done by the gas along path $A \to B \to C$. Then the process $A \to B$ is:

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$A$ monoatomic gas $(\gamma = 5/3)$ at a temperature of $27^{\circ}C$ is compressed adiabatically to $8/27$ of its initial volume. The change in temperature is . . . . . . $K$.

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In a thermodynamic process,if $\Delta W$ and $\Delta U$ are the work done and the change in the internal energy of a system respectively,then:

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$\Delta U + \Delta W = 0$ is valid for

The pressure of a motorcar tire at $27^{\circ}C$ is $2$ atmospheres. If the tire suddenly bursts,the final temperature will be ....... $K$ (Take $\gamma = 1.4$).

$A$ fixed amount of a gas undergoes a thermodynamic process as shown in the figure,such that the heat interaction along path $B \to C \to A$ is equal to the work done by the gas along path $A \to B \to C$. Then the process $A \to B$ is:

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$A$ monoatomic gas $(\gamma = 5/3)$ at a temperature of $27^{\circ}C$ is compressed adiabatically to $8/27$ of its initial volume. The change in temperature is . . . . . . $K$.

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