Phy202
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Prof. Smith, can you help me with the following?
*For the thermal conductivity Q/t how can I covert this to Watts?
Thank you
If Q is the amount of thermal energy flowing in time t, then the SI units of Q would be Joules. The SI units of t would be seconds. The units of Q / t would be Joules / second, or watts.
If Q is in some other unit it can be converted to Joules.
If t is in some other unit, it can easily be converted to seconds.
*How is thermal energy, internal energy, and work required found after solving for T2 degrees in Kelvin? I was using the formula (P1*V1)/T1(P2*V2)/T2 to solve for T2.
The answers would depend on how you get from State 1 to State 2.
If the change occurs at constant volume, then the thermal energy required is n * C_v * `dT. The P vs. V graph is a vertical line segment, and the area beneath this segment is 0. The area beneath a P vs. V graph is equal to the work done, so no work is done by or on the system. All the thermal energy going into the system goes into internal energy.
If the change occurs at constant pressure, then the thermal energy required is n * C_p * `dT. The P vs. V graph is a horizontal line segment, and the area beneath the segment is P * `dV, which is the work done by the system. The thermal energy that goes into the system does work and changes the system's internal energy, and by energy conservation this accounts for all the energy. Thus `dQ = `dW_BY + `dU, where `dQ is the thermal energy going into the system, `dW the work done by the system, and `dU the change in the system's interval energy.
If the expansion or compression is adiabatic then the situation is more difficult to analyze, because the temperature, pressure and volume all change in such a way that no thermal energy is exchanged. The area beneath the P vs. V graph is significant, and the corresponding work is done at the expense of the internal energy of the gas. Thus in an an adiabatic expansion,
`dQ = 0 so
`dW_by + `dU = 0 (work done by the gas is equal and opposite to the change in internal energy).
If you have additional questions, or if you want to apply these ideas to a specific situation and send me your work, I'll be glad to comment further.