Heat transfer to an ideal gas [closed]








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In thermodynamics, why is the heat that is transferred to an ideal gas considered as work that was done on the system for an isothermal process?




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In general, heat transferred to an ideal gas is *not* considered as work that was done on the system. I suspect you're referring to a process in which the internal energy doesn't change: $\Delta{U}=0$. As $\Delta{U} = q + w$ (where q is heat and w is work), in the case where $\Delta{U}=0$ you get $q=-w$. This holds for a general system (not just ideal gases).
For an ideal gas, U only depends on temperature. So for an isothermal process, the internal energy of the ideal gas will not change, and you'll once again get the result $q=-w$. 


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