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Ruby_338
- 27
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How come ##adiabatic \,work = C_v(T_2 -T_1)##
Maybe he means "per mole" or "per unit mass.". Also, irrespective of the 2nd law, Q is equal to zero for an adiabatic process.Ssnow said:If the transformation is adiabatic you have that ##\Delta U=-W## for the second law of thermodynamics because ##Q=0##. So ##W=-\Delta U = n\cdot C_{v}(T_{1} -T_{2})## where ## C_{v}## is the specific heat for constant volume and ##n## represents the mole. It is strange that in your formula there isn't ##n##...
Ssnow
In an adiabatic process, work done is defined as the energy transferred to or from a system through mechanical means, such as compression or expansion. This can also be thought of as the force applied over a distance.
The equation for calculating work done in an adiabatic process is W = -PΔV, where W is work done, P is pressure, and ΔV is the change in volume.
In an adiabatic process, there is no transfer of heat, so all the work done is due to changes in pressure and volume. This is different from other thermodynamic processes, such as isothermal or isobaric, where there may be heat transfer and work done simultaneously.
Yes, the work done in an adiabatic process can be negative. If the system is expanding, work is being done by the system and the work done will be negative. On the other hand, if the system is being compressed, work is being done on the system and the work done will be positive.
The first law of thermodynamics 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. In an adiabatic process, there is no heat transfer, so the change in internal energy is solely due to the work done. This means that the work done in an adiabatic process is equal to the change in internal energy of the system.