
Which among the following is a feature of adiabatic expansion?
A. $\Delta V < 0$
B. $\Delta U < 0$
C. $\Delta U > 0$
D. $\Delta T = 0$
Answer
563.1k+ views
Hint: We know that an adiabatic process occurs without transporting heat or mass between a thermodynamic system and its surroundings. It supports the theory that is employed to explain the first law of thermodynamics and is a major thermodynamic concept.
Complete step by step answer:
For an adiabatic free expansion of an ideal gas, the gas is kept in a container that is insulated and then allowable to increase in a vacuum. As there is no external pressure for the gas to increase against, the work done by or on the system takes the value zero. Since this process does not include any interchange of transfer of heat or work, the first law of thermodynamics states that the net internal energy change of the system is zero. For an ideal gas, the temperature remains the same since the internal energy is dependent on temperature in that case.
For a free expansion, the work done would be zero.
$W = 0$
For an adiabatic process, the heat would be zero.
$q = 0$
So, from the first law of thermodynamics, the change in internal energy is the sum of total heat and the work done.
$\Delta U = q + w$
Substituting the values of q and w, we would get
$\Delta U = q + w$
$\Delta U = 0 + 0$
$\Delta U = 0$
From the equation, we could observe that there is no change in the internal energy, so the temperature would also remain fixed.
The equation is,
$\Delta T = 0$
So, $W = 0$, $q = 0$ and $\Delta T = 0$ for free expansion for gas that is ideal in adiabatic condition.
Therefore, the option D is correct.
Note: We have to remember that when the pressure of a gas is improved by work done on it by its surroundings, the process of adiabatic heating occurs. In the atmosphere of Earth when an air mass descends, the process of adiabatic heating happens. Adiabatic cooling occurs when the pressure on an adiabatically isolated system is decreased, allowing it to enlarge, thus leading it to accomplish work on its surroundings.
Complete step by step answer:
For an adiabatic free expansion of an ideal gas, the gas is kept in a container that is insulated and then allowable to increase in a vacuum. As there is no external pressure for the gas to increase against, the work done by or on the system takes the value zero. Since this process does not include any interchange of transfer of heat or work, the first law of thermodynamics states that the net internal energy change of the system is zero. For an ideal gas, the temperature remains the same since the internal energy is dependent on temperature in that case.
For a free expansion, the work done would be zero.
$W = 0$
For an adiabatic process, the heat would be zero.
$q = 0$
So, from the first law of thermodynamics, the change in internal energy is the sum of total heat and the work done.
$\Delta U = q + w$
Substituting the values of q and w, we would get
$\Delta U = q + w$
$\Delta U = 0 + 0$
$\Delta U = 0$
From the equation, we could observe that there is no change in the internal energy, so the temperature would also remain fixed.
The equation is,
$\Delta T = 0$
So, $W = 0$, $q = 0$ and $\Delta T = 0$ for free expansion for gas that is ideal in adiabatic condition.
Therefore, the option D is correct.
Note: We have to remember that when the pressure of a gas is improved by work done on it by its surroundings, the process of adiabatic heating occurs. In the atmosphere of Earth when an air mass descends, the process of adiabatic heating happens. Adiabatic cooling occurs when the pressure on an adiabatically isolated system is decreased, allowing it to enlarge, thus leading it to accomplish work on its surroundings.
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