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Previously, we said that the enthalpy of an ideal gas is independent of pressure at constant temperature.
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And the internal energy of an ideal gas is independent of volume at constant temperature.
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The enthalpy and internal energy of an ideal gas were asserted to be functions of temperature only.
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Here, we can prove it using the property relations. Let's begin with enthalpy. dH is TdS + VdP.
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Divide both side by dP at constant temperature. Then (dH over dP) is T times (dS over dP) + V.
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So, to get the pressure dependence of enthalpy at constant temperature,
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we need (dS over dP) at constant temperature. Among the property relations, it can be obtained from dG,
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since G has variables of P and T. (dS over dP) at constant T is - (dV over dT) at constant P. So,
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(dH over dP) is now - T (dV over dT) + V. For ideal gas, V equals to RT over P from the equation of state.
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Then, (dH over dP) at constant T becomes zero.
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Thus, enthalpy does not depend on pressure at constant T and it is a function of temperature only.
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Similarly, let's prove that the internal energy of an ideal gas is a function of temperature only
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and independent of volume. dU is TdS - PdV. Divide both sides with dV at constant T.
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Then the volume dependence of the internal energy can be calculated from (dS over dV) at constant T.
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This can be obtained from dF since F has T, V as variables. dF is - SdT - PdV.
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The property relation from here is (dS over dV) at constant T equals (dP over dT) at constant V.
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So, (dU over dV) is T times (dP over dV) - P. Again P is RT over V for the ideal gas.
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Thus (dU over dV) at constant T is zero. So the internal energy is a function of temperature only.