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# During an isothermal expansion of an ideal gas, its:A. Internal energy decreaseB. Enthalpy decreasesC. Enthalpy remains unaffectedD. Enthalpy reduce to zero

Last updated date: 20th Jun 2024
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Hint: The isothermal expansion means a gas is expanding from initial volume to final volume at constant temperature. The internal energy and enthalpy depend upon the temperature only in case of ideal gas.

Complete Step by step answer: The thermodynamic process in which the temperature remains constant throughout the process is known as the isothermal process. The process in which the volume of the thermodynamic system increases is known as expansion. So, during the isothermal expansion of an ideal gas, the temperature remains constant and volume increases.
Internal energy is a function of temperature and volume.
${\text{U = }}\,{\text{f}}\,\left( {{\text{T,V}}} \right)$
Where,
U is the internal energy.
T is the temperature.
V is the volume.
${\text{dU = }}\,{\left( {\dfrac{{{\text{dU}}}}{{{\text{dT}}}}} \right)_{\text{V}}}{\text{dT + }}{\left( {\dfrac{{{\text{dU}}}}{{{\text{dV}}}}} \right)_{\text{T}}}{\text{dV}}$…..$(1)$
According to Joule’s law, the value of change in internal energy with respect to volume change is zero. So,
${\left( {\dfrac{{{\text{dU}}}}{{{\text{dV}}}}} \right)_{\text{T}}} = 0$
So, the equation $(1)$can be reduced as follows:
${\text{dU = }}\,{\left( {\dfrac{{{\text{dU}}}}{{{\text{dT}}}}} \right)_{\text{V}}}{\text{dT}}$
In isothermal expansion, temperature remains constant so, the change in internal energy is zero means internal energy does not get affected.
Enthalpy and internal energy relation is as follows:
${\text{H}}\,{\text{ = }}\,{\text{U}}\,{\text{ + }}\,{\text{pV}}$
Where,
H is the enthalpy.
P is the pressure.
According to ideal gas,
${\text{pV = }}\,{\text{nRT}}$
So,
${\text{H}}\,{\text{ = }}\,{\text{U}}\,{\text{ + }}\,{\text{nRT}}$
Internal energy is a function of temperature and pressure and volume product is also a function of temperature so, enthalpy is the function of temperature in the case of ideal gas so, the change in enthalpy is also zero because the temperature is constant. So, enthalpy also remains unaffected during the isothermal expansion process.
So, enthalpy and internal energy both remain unaffected in the isothermal expansion process.

Therefore, option (C) enthalpy remains unaffected, is correct.

Note: In the isothermal process, change in enthalpy and internal energy is zero. Work depends upon the volume change. The formula of work done in the isothermal expansion process is, ${\text{w = nRT}}\,{\text{ln}}\dfrac{{{{\text{V}}_2}}}{{{{\text{V}}_1}}}$ .
According to the first law of thermodynamics,$\Delta {\text{U}}\,{\text{ = }}\,{\text{q}}\,{\text{ + w}}$ . Change in internal energy is zero, so, . the formula to calculate the heat change in the isothermal process is, ${\text{q = }}\,{\text{ - }}\,{\text{nRT}}\,{\text{ln}}\dfrac{{{{\text{V}}_2}}}{{{{\text{V}}_1}}}$.