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Adiabatic process-how is internal energy mcv(t2-t1)

• Thread starter Thread starter sreeram
• Start date Start date Nov 14, 2009
• Tags Tags Adiabatic Energy Internal Internal energy

Click For Summary In a polytropic process involving an ideal gas, the change in internal energy is expressed as ΔU = mC_VΔT, which holds true regardless of whether the process is at constant volume or constant pressure. This relationship arises from the unique property of ideal gases, where internal energy depends solely on temperature, leading to the equation ΔU = m(C_P - R)ΔT as well. The constants C_V and C_P do not impose constraints on the process type, making them applicable across various thermodynamic processes. The derivation of this relation is based on the assumption of negligible interactions between gas particles, resulting in energy being a function of temperature alone. Understanding these principles can be challenging for thermodynamics students but is essential for grasping the behavior of ideal gases. sreeram Messages 10 Reaction score 0 In polytropic process internal energy change is=mCv(T2-T1)? why is it not mCp(T2-T1)? Why are we not taking specific heat at constant pressure? Last edited: Nov 15, 2009 Physics news on Phys.org

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Mapes Science Advisor Homework Helper Gold Member Messages 2,591 Reaction score 21 Are you dealing with an ideal gas? If so, the internal energy is U=mC_VT+U_0=m(C_P-R)T+U_0 (the internal energy depends only on temperature), so \Delta U=mC_V\Delta T=m(C_P-R)\Delta T. C_V (or C_P-R) are just constants here. The equations apply to any process. sreeram Messages 10 Reaction score 0 mCv(t2-t1) is the heat required to raise the temperature of mass 'm' from t1 to t2 at constant volume. But adiabatic process is not a constant volume process. My question is how does this equation give the value of change in internal energy? Mapes Science Advisor Homework Helper Gold Member Messages 2,591 Reaction score 21 Are you dealing with an ideal gas? sreeram Messages 10 Reaction score 0 yes. I am dealing with ideal gas Mapes Science Advisor Homework Helper Gold Member Messages 2,591 Reaction score 21 It's a unique property of ideal gases that \Delta U=mC_V\Delta T holds for any process. The constant-volume constraint is not required. That's why I also wrote the equation as \Delta U=m(C_P-R)\Delta T, which similarly doesn't mean that constant pressure is required just because the constant-pressure specific heat appears. Again, C_V and C_P are just acting as constants, and do not represent constraints on the equation. It's hard for every thermo student to get used to this, but that's how it is. sreeram Messages 10 Reaction score 0 Thanks for the answer. But how did they found this relation? Mapes Science Advisor Homework Helper Gold Member Messages 2,591 Reaction score 21 It comes from one of the assumptions we make for an ideal gas, that there's no interaction between the atoms or molecules. If this is true, then changing the pressure (while keeping temperature constant) shouldn't have any effect on the total energy (i.e., (\partial U/\partial P)_T=0). We're left with a energy dependence on temperature alone. Gerenuk Messages 1,027 Reaction score 5 Starting from <br /> \mathrm{d}U=T\mathrm{d}S-p\mathrm{d}V<br /> one can in fact show that for the special case <br /> p=Tf(V)<br /> i.e. pressure proportional to temperature at constant volume, the energy is a function of temperature only U=A(T). The proportionality to temperature doesn't follow from it yet, but if you also assume that the density of states (statmech treatment) is a power law, the you can show that U=CT.

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