Is it theoretically possible to keep the volume and temperature of a gas undergoing a chemical reaction, to which heat has been supplied, constant? I was reading about enthalpy$^*$ when I thought that if it is possible for the gases undergoing a reaction to be supplied heat while keeping the pressure and temperature constant and just allowing it's volume to change, is it theoretically possible to supply heat to gases undergoing a reaction by keeping the temperature and volume constant and allowing only the pressure to change?
My intuition says that this is not possible, since pressure is the result of molecule collision. On heating, the collisions get more vigorous hereby increasing the pressure. But the increased molecular vibrations would cause a rise in temperature. I am not sure how correct my intuition is and would like someone more knowledgeable than me to weight in.
My question is not a duplicate and nor related to this question
$* $: In my book, enthalpy is defined as:

heat evolved or absorbed in a reaction at constant temperature and constant pressure

 A: It is neither possible to change the pressure of a gas while keeping the volume and temperature constant, nor to change the volume whilst keeping the pressure and temperature constant. This can easily be seen from the ideal gas law.
$$
p = \frac{Nk_B T}{V}
$$
If the temperature and volume are constant the left hand side of the equation is constant and so the pressure must be constant. We can use a similar argument with a simple rearrangement of the equation for the case of constant pressure and temperature.
The same argument applies to non-ideal gases, although rearranging the equation may be more complex. The equation of state will quite generally give a relation between temperature, pressure and volume. You can always, in principle, solve this for one in terms of the others, so fixing any two parameter will determine the third.
There is a caveats to this. I have assumed that the amount of gas is fixed. If this is not true then we can, for example, change the pressure at constant volume and temperature by changing the amount of gas. Notable cases where this might actually occur are when the gas is undergoing a chemical reaction or a phase transition
A: By "heat" a gas at constant temperature, I assume you mean increase the internal energy of the gas.  For an ideal gas, internal energy is a function of temperature only, so the internal energy can only be increased by increasing the temperature.
For a real (non-ideal) gas, internal energy also depends slightly on pressure, but this is typically a small effect. (For example, in Joule, or free, expansion there is a small decrease in gas temperature with decrease in pressure.)
It may be possible to slightly increase the internal energy by increasing the pressure with fixed temperature and volume.
If the number of moles of the gas are allowed to change (mass transfer), at constant volume the temperature would have to decrease to maintain constant pressure.
Regarding your edited question.  For a fixed volume, if the gas underdoes an endothermic reaction that decreases temperature but increases the number of moles, then heat is added to restore the original temperature, the pressure could increase due to the increased number of moles.
(Also, the actual definition for Enthalpy is internal energy plus pressure times volume: $H = U + PV$.)
A: The ideal gas law is
P V = n R T
P = Pressure
V = Volume
n = quantity of gas
R = a constant based on the type(s) of gas
T = Temperature
You want to hold T and V constant while changing P
That is only possible if n changes in proportion to P (or to be more precise, n R changes in proportion to P).

*

*change n by adding or removing gas

*change R ... which is generally not that easy.

A: As you might know the very first equation of thermodynamics states that some part of heat provided to gas goes to change in internal energy of gas and other to perform work..
As stated by you already that
neither temperature can change( which implies no change in internal energy  )
$\Delta$ U =0
and W=0 so NO HEAT EXCHANGE IS POSSIBLE
Q=$\Delta$U +W
