# I urgently need help with thermo, closed systems please

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#1
Hello everyone.

I am revising for thermo but I am finding myself strugling to understand everything when it comes to closed systems as everything is a bit mixed in my head at the minute.

Basically, for closed systems we have the formulas:

Non-flow Energy Equation: U2-U1 = Q - W

Non-flow, No phase change, Constant c; U2-U1 = mct

Non-flow, With phase change, Constant c; U2-U1 = mc1(t1-t0) + mcphaset1 + mc2(t2-t1)

Could someone explain to me which is the different on these three processes and also how are these three processes realated Isochoric processes, isobaric, isothermal and adiabatic.
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1 year ago
#2
(Original post by LaurenWilliamss)
Hello everyone.

I am revising for thermo but I am finding myself strugling to understand everything when it comes to closed systems as everything is a bit mixed in my head at the minute.

Basically, for closed systems we have the formulas:

Non-flow Energy Equation: U2-U1 = Q - W

Non-flow, No phase change, Constant c; U2-U1 = mct

Non-flow, With phase change, Constant c; U2-U1 = mc1(t1-t0) + mcphaset1 + mc2(t2-t1)

Could someone explain to me which is the different on these three processes and also how are these three processes realated Isochoric processes, isobaric, isothermal and adiabatic.

…. Basically, for closed systems we have the formulas:

Non-flow Energy Equation: U2-U1 = Q - W

Non-flow, No phase change, Constant c; U2-U1 = mct

Non-flow, With phase change, Constant c; U2-U1 = mc1(t1-t0) + mcphaset1 + mc2(t2-t1)

Could someone explain to me which is the different on these three processes …
TBH, I am not sure what are you asking here. What is the system that you are considering here – ideal gas or real gas or solid or liquid? Which processes are you referring to?

… Isochoric processes, isobaric, isothermal and adiabatic.
For isochoric process of an ideal gas system, work done on the system is zero. The first law of thermodynamics becomes:

Change in internal energy = Heat gained or lost

ΔU = Q

For isobaric process of an ideal gas system, work done on the system is just

W = ‒p(V2V1)

The first law of thermodynamics becomes
ΔU = Q + W

ΔU = Q + [‒p(V2V1)]

For isothermal process of an ideal gas system, the temperature remains unchanged, so the change in internal energy is zero. The first law of thermodynamics becomes
0 = Q + W

For adiabatic process of an ideal gas system, no heat transfers into or out of a system; Q = 0. The first law of thermodynamics becomes
ΔU = W
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