internal energy ideal gas

[905290504]

why is the ∆U = 0 for an isothermal expansion of an ideal gas?

[Edison Tham 3D]

For an isothermal expansion of an ideal gas, the amount of work that the system does is replaced by the same amount of heat (which is how the system is isothermal). Since we know that ∆U=q+w, the work done to the system is going to be negative (for expansion) and the heat would be put into the system (the same value but positive), so overall ∆U would equal 0.

[Alexandra Salata 2L]

In Isothermal process the temperature is constant. The internal energy is a state function dependent on temperature. Therefore, the internal energy change is 0.

[Chinmayi Mutyala 3H]

The work done is replaced by heat so Q = -W. Since delta U = Q + W, it would be 0.

[Andrew Yoon 3L]

∆U = 0 for an isothermal expansion of an ideal gas because the work that is done within the system is replaced by energy in the form of heat. This means that Q = -W, which also can be written as ∆U = Q + W, which also equals 0.

[Susanna Givan 2B]

Are there other instances in which Delta U is 0?