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(C) For an ideal gas,the internal energy $ \Delta U $ and enthalpy $ \Delta H $ are functions of temperature only. For an isothermal process,the chang

Vedclass · Accepted Answer

$ \Delta H = 0 $ and $ \Delta U = 0 $

Vedclass · Answer

(C) For an ideal gas,the internal energy $ \Delta U $ and enthalpy $ \Delta H $ are functions of temperature only. For an isothermal process,the change in temperature $ \Delta T = 0 $. Since $ \Delta U = n C_V \Delta T $,if $ \Delta T = 0 $,then $ \Delta U = 0 $. Similarly,since $ \Delta H = \Delta U + n R \Delta T $,if $ \Delta T = 0 $ and $ \Delta U = 0 $,then $ \Delta H = 0 $. Therefore,for an isothermal reversible expansion of an ideal gas,both $ \Delta H = 0 $ and $ \Delta U = 0 $.

For the isothermal reversible expansion of an ideal gas,which of the following is correct?

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