Is heat added during an isothermal process zero? I am confused about the first law of thermodynamics when applied to an isothermal process for an ideal gas. In my textbook it says that for any process involving an ideal gas:
$dU=dQ=nC_V∆T$
Then doesn’t this imply that since temperature doesn’t change, there is not heat added either. However, if $∆U = 0$, then $Q = W$  for an ideal gas and consequently $Q$ cannot equal zero. I am not sure whether I have gotten something conceptually wrong as intuitively I just feel that since the temperature doesn’t change during an isothermal process heat is not added to the system.
 A: The ideal gas law says that for a fixed number of molecules in a gas
$$\frac{PV}{T}=\text{constant}.$$
If the process is isothermal, as you specify, then $T$ is constant and we can write
$${PV}=\text{constant}.$$
This means that in an isothermal process, both pressure and volume must change in order for any change in state to occur.
Now, you mention the equation $$dU=dQ=nC_V∆T.$$
The subscript on the specific heat value, $C_V$, means that this applies for constant volume. So you have a situation in which this calculation does not apply. Also, notice that if $V$ is constant, then
$$\frac{P}{T}=\text{constant}$$
and no work is done, because work requires a change in volume. This is incompatible with an isothermal process.
What you should realize is that this is not the only method of calculating heat addition or removal. Also, the first law says
$$\text{d}U=Q_{in}+\text{d}W$$
In this particular form of the statement, $Q_{in}$ positive if heat flows into a system and negative if heat is removed from the system, and $\text{d}W$ is positive if work is done on the system and negative if work is done by the system.
In an isothermal process, with both $P$ and $V$ changing, but $T$ being constant, work is being done on (or by) the system so the internal energy must change unless heat is removed (or added to) the system, respectively.
In thermodynamic processes you must be careful to understand when you are applying a specialized condition (heat in a constant volume situation, which is incompatible with isothermal processes) rather than a general condition (first law of thermodynamics).
A: 
$dU=dQ=nC_V∆T$

Does your book really state that $Q=nC_{V}\Delta T$ for an isothermal process involving an ideal gas? Because, as far as I know, that would apply to an isochoric (constant volume) process.
On the other hand, it can be shown that
$$\Delta U=nC_{V}\Delta T$$
for any process (not just constant volume) involving an ideal gas.
For a reversible isothermal ($\Delta T=0$) expansion process of an ideal gas  the first law says the heat added $Q$ (energy in) exactly equals the work $W$ done (energy out) by the gas, for $\Delta U=0$.
The work done by the gas is
$$W=nRTln\frac{v_2}{v_1}=nRTln\frac{P_1}{P_2}$$
and that equals $Q$.
Hope this helps.
A: I think you are confused about $\Delta Q$ and $\Delta H$ ,  during isothermal process enthalpy change i.e $\Delta H$ = $0$ but $\Delta Q$ is not zero, it is only because enthalpy change is a function of temperature but ,in isothermal process temperature is constant .
Enthalpy is heat absorbed at constant pressure, while $\Delta Q$ is heat exchange for any process (for adiabatic process  $Q$ = $0$ as well as $\Delta H$)
Now you, see if a gas is undergoing isothermal process it's temperature shall remain constant which means at any instant-  $$P V = nRT= constant \tag{1} $$
now we know $$\Delta H = \Delta U + P V \tag{2}$$
you can clearly see that the $P V$ term is constant  and we also know $\Delta U$ is also function of temperature so internal energy is constant as well which makes the whole enthalpy change in isothermal process zero.
But we all know that in isothermal process  some work is done, so it is the heat exchange i.e $\Delta Q$  which gets consumed in isothermal work done
Conclusion- In isothermal process $\Delta H=0$, $\Delta U=0$ but $\Delta Q$ is not $0$ . and also $\Delta Q$ is path dependent function not temperature dependent like enthalpy and internal energy.
A: In thermodynamics, we no longer define heat capacity in terms of the amount of heat flow Q.  Instead, because heat capacity is supposed to be a physical property of the material, rather than a feature of any specific process, we define heat capacity in terms of two key physical properties of state, internal energy U and enthalpy H, as follows:  $$nC_v\equiv\left(\frac{\partial U}{\partial T}\right)_V$$
$$nC_p\equiv\left(\frac{\partial H}{\partial T}\right)_P$$Once you accept and internalize these two precise mathematical definitions of the heat capacities at constant volume and at constant pressure, respectively, all you confusion should vanish.
A: Since two of my friend use a formula based approach so my focus would be on logical approach, consider a gas that is allowed to slowly escape from a container immersed in a constant-temperature bath. As the gas expands, it does work on the surroundings and therefore tends to cool, but the thermal gradient that results causes heat to pass into the gas from the surroundings to exactly compensate for this change. This is called an isothermal expansion. In an isothermal process the internal energy remains constant and we can write the First Law as 0 = q + w, or q = –w, illustrating that the heat flow and work done exactly balance each other.
Because no thermal insulation is perfect, truly adiabatic processes do not occur. However, heat flow does take time, so a compression or expansion that occurs more rapidly than thermal equilibration can be considred adiabatic for practical purposes.
If you have ever used a hand pump to inflate a bicycle tire, you may have noticed that the bottom of the pump barrel can get quite warm. Although a small part of this warming may be due to friction, it is mostly a result of the work you (the surroundings) are doing on the system (the gas.)
I have used the equation in chemistry so change sign according to you))
