I have a confusion when you explained Spring piston work: You wrote that Pressure increases linearly with V why is that so, because pressure decreases with increase in volume? which pressure are you talking about? Is that the pressure exerted by the gas or the pressure exerted by the spring piston?
Because of equilibrium, the pressure exerted by the spring IS equal to the pressure exerted by the gas. One is pushing out, one is pushing in. And the force of the spring is F=kx so it is increasing linearly with x, so the pressure (since area is constant) is increasing linearly with x (and V).
For a constant volume process would it not matter if temperature changes? such as a rigid body container problem, in which you have 2 states at different tempts, changing pressures, and therefore constant volume? Its still 0?
XplosiveAction correct. For a constant volume process the boundary work would be zero, even though temperature and pressure are probably changing. If the volume (and therefore the “boundary”) is not changing then the boundary work is zero.
Engineering Deciphered thanks for the very quick reply I appreciate it greatly. I was asking because specifically I’m working on a problem right now which only states: “A rigid body contains superheated water vapor at 2.5MPa and 800 deg.Celsius. The system is cooled until the water is at 500kPa and 500 deg.Celsius. Calculate the boundary work done in kJ.” And the only thing that I learned was that the Energy Balance and Rate Balance equations made use of internal energy rather than enthalpy in those equations for a rigid body, but nothing about how to go about a rigid body problem since that was the only exception for the topic I’m learning.
XplosiveAction you’re right about internal energy and enthalpy. Generally when I say “rigid tank” in a problem I want that to mean constant volume. That’s what I’d assume “rigid body” means, but other teachers might describe it differently.
It’s both Pv=RT or PV=mRT because v=V/m. Lowercase v is specific volume, uppercase V is volume. The units will help you out- units have to be the same on both sides of the equation. I never use nRT.
@@engineeringdeciphered n=mass/molarmass which makes sense and is consistent with what you've been teaching. R=will be the universal gas constant(not for a specific substance).
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this help me a lot thanks
Im struggling so much thinking where the ln suddenly came from in the isothermal compression of ideal gas eq, this really helps.
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in 16:40 we should multiply with the same term the whole expression either p1v1 or p2v2
Thank a lot !!!!
I have a confusion when you explained Spring piston work:
You wrote that Pressure increases linearly with V why is that so, because pressure decreases with increase in volume?
which pressure are you talking about?
Is that the pressure exerted by the gas or the pressure exerted by the spring piston?
Because of equilibrium, the pressure exerted by the spring IS equal to the pressure exerted by the gas. One is pushing out, one is pushing in. And the force of the spring is F=kx so it is increasing linearly with x, so the pressure (since area is constant) is increasing linearly with x (and V).
@@engineeringdeciphered thanks sir :)
ty
عاشت ايدك ❤️😞
انتي الي فبالي؟🙄
does anyone know if when he says expansion = work out and compression = work in at 3:54, is it for reversible work?
Sir will you explain how you got the last question?😢
For a constant volume process would it not matter if temperature changes? such as a rigid body container problem, in which you have 2 states at different tempts, changing pressures, and therefore constant volume? Its still 0?
XplosiveAction correct. For a constant volume process the boundary work would be zero, even though temperature and pressure are probably changing. If the volume (and therefore the “boundary”) is not changing then the boundary work is zero.
Engineering Deciphered thanks for the very quick reply I appreciate it greatly. I was asking because specifically I’m working on a problem right now which only states:
“A rigid body contains superheated water vapor at 2.5MPa and 800 deg.Celsius. The system is cooled until the water is at 500kPa and 500 deg.Celsius. Calculate the boundary work done in kJ.”
And the only thing that I learned was that the Energy Balance and Rate Balance equations made use of internal energy rather than enthalpy in those equations for a rigid body, but nothing about how to go about a rigid body problem since that was the only exception for the topic I’m learning.
XplosiveAction you’re right about internal energy and enthalpy. Generally when I say “rigid tank” in a problem I want that to mean constant volume. That’s what I’d assume “rigid body” means, but other teachers might describe it differently.
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Im always confused is it PV = RT or PV = mRT or PV = nRT
It’s both Pv=RT or PV=mRT because v=V/m. Lowercase v is specific volume, uppercase V is volume. The units will help you out- units have to be the same on both sides of the equation. I never use nRT.
@Engineering Deciphered Thank you so much. Ya I went back and understand it now. Again thank you so much for the quick reply. You are a legend.
@@engineeringdeciphered n=mass/molarmass which makes sense and is consistent with what you've been teaching. R=will be the universal gas constant(not for a specific substance).