
For the isothermal expansion of an ideal gas:
A. U and H increases.
B. U increases but H decreases
C. H increases but U decreases.
D. H and U are unaltered
Answer
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Hint:An isothermal process is a type of thermodynamic process in which the temperature of the system remains constant. In an isothermal expansion Enthalpy (H) is the sum of the internal energy (U) and the product of pressure and volume (PV).
Complete answer:
In an ideal gas, the amount of energy entering the environment is equal to the work done on the gas, hence no change is observed in internal energy $(\Delta U = 0)$ .
In an ideal gas the internal energy U only depends on the amount of substance n and temperature T, and in a closed system n is constant. $(\Delta n = 0)$.
During an isothermal process temperature T remains constant $(\Delta T = 0)$.
Enthalpy H is defined as $H = U + PV$ and the ideal gas law states that $PV = nRT$.
Thus $H = U + nRT$
Since it is a closed system and Temperature is constant due to isothermal process the product nRT is constant, and therefore, according to the ideal gas law, also the product PV is constant. Furthermore, since U is constant during the given process, the sum $H = U + nRT$ has to remain unchanged.
Therefore, no change is observed in the enthalpy and internal energy in the case of ideal gas.
Hence the correct option is option (D). Both remain unaltered.
Note:
In an Isothermal Expansion Temperature is held constant, therefore the change in energy is zero (U=0). So, the heat absorbed by the gas equals the work done by the ideal gas on its surroundings. Enthalpy change is also equal to zero because the change in energy zero and the pressure and volume is constant.
Complete answer:
In an ideal gas, the amount of energy entering the environment is equal to the work done on the gas, hence no change is observed in internal energy $(\Delta U = 0)$ .
In an ideal gas the internal energy U only depends on the amount of substance n and temperature T, and in a closed system n is constant. $(\Delta n = 0)$.
During an isothermal process temperature T remains constant $(\Delta T = 0)$.
Enthalpy H is defined as $H = U + PV$ and the ideal gas law states that $PV = nRT$.
Thus $H = U + nRT$
Since it is a closed system and Temperature is constant due to isothermal process the product nRT is constant, and therefore, according to the ideal gas law, also the product PV is constant. Furthermore, since U is constant during the given process, the sum $H = U + nRT$ has to remain unchanged.
Therefore, no change is observed in the enthalpy and internal energy in the case of ideal gas.
Hence the correct option is option (D). Both remain unaltered.
Note:
In an Isothermal Expansion Temperature is held constant, therefore the change in energy is zero (U=0). So, the heat absorbed by the gas equals the work done by the ideal gas on its surroundings. Enthalpy change is also equal to zero because the change in energy zero and the pressure and volume is constant.
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