As we don't consider molecular volume or any interaction between molecules in an ideal gas, what causes its volume? What is the volume given by PV=nRT?
If we take a balloon as an example, what restricts it from being shrunk? I know it is due to pressure. But I want to know about it in the atomic scale. What stops the molecules from getting much closer? Is it connected to their Kinetic energy?
The volume of an ideal gas is simply the volume of its container. That's because one of the assumptions for a gas to be considered ideal is that the total volume of the individual molecules is magnitudes smaller than the volume that the gas occupies.
Hope this helps.
In the ideal gas law $pV = nRT$, the $V$ you find is the volume of the container of the gas, $p$ as the pressure the gas has against the border of the container (in your example, that's the plastic layer of the baloon).
Let's have a look inside the gas. On one hand you have the collision of the particle against the surface. In each path (one way and return) of lenght $L$, a molecule exchange momentum from $+mv$ ($m$ is the mass of the molecule, $v$ its speed) to $-mv$ in one way and from $-mv$ to $+mv$ in the return. Consider the surface, this means $2mv$ momentum exchange in the time the molecule makes 2L that is $t = \frac{2L}{v}$. Totally, a surface "feels" one molecule for a $\Delta p = 2mv$ momentum exchange in $t = 2L/v$ period, that is a net force: $$ F = \frac{\Delta p}{t} = \frac{2mv}{2L/v} = m v^2 \frac{1}{L} $$ Extend this behaviour to all N molecules: $$ F_\text{tot} = \sum^N F = N m v^2 \frac{1}{L} $$ The pressure is defined as the ortogonal component of the force respect of the surface S where it acts, so: $$ p = \frac{F_\text{tot}}{S} = N m v^2 \frac{1}{LS} = \frac{Nmv^2}{V} $$ where we put $SL = V$ for a cubic container but this is valid also to other containers knowing that you can decompose a volume in cubes (like what we do in a Riemann integral in 2 dimensions). What is the speed $v$ just inserted? The $v^2$ speed seen in that equation should be the average $v_i^2$ for all the molecules in the direction ortogonal to the surface, so $\langle v_x^2\rangle$. Because of isotropy of the container and space, this is only a third part of the medium squared velocity: $$ \langle v_x^2 \rangle = \frac{1}{3} \langle v^2 \rangle \equiv \frac{1}{3} v_\text{RMS}^2 $$ This $v_\text{RMS}^2$ speed comes from the kinetic energy shared among them, while they are in thermal equilibrium (indeed you have a definite $T$ for your system). The equipartition principle states that for each degree of freedom, the associated energy ($\frac{1}{2}mv_i^2$ in the $i$ direction for the kinetic energy related to one degree of freedom) is equal to $\frac{1}{2}k_\text{B}T$, so having three degree of spatial traslation for monoatomic gas means: $$ 3\cdot \frac{1}{2}k_\text{B}T = \frac{1}{2}m (v^2_x + v_y^2 + v_z^2) \equiv \frac{1}{2} m v^2_\text{RMS}$$ where $v_\text{RMS}$ is the root-mean-squared speed, something that expresses the temperature of the ideal gas, as you can see in the above equation. Coming back to the pressure, with $v_\text{RMS}^2$: $$ p = \frac{Nmv^2_\text{RMS}}{3V} = \frac{N}{V} \frac{2}{3} \frac{1}{2} m v_\text{RMS}^2 = \frac{1}{V} \frac{2}{3} U = \frac{2}{3} \frac{U}{V} $$ So you find: $$ pV = \frac{2}{3}U = \frac{2}{3} N\frac{3}{2} k_\text{B}T = n R T$$ It is correct to say that molecules don't interact each other, but it is not correct to say that molecules don't interact at all, because to be confined in the container (in your example, the baloon) they need to exchange momentum with the borders they bump into in each trajectory. Exchanging momentum means that in every scattering the component of the momentum that is ortogonal to the surface must change sign keeping the same direction.
As you note, the particles in the ideal gas are assumed to have no volume and to not interact. How, then, does the ideal gas maintain a volume or resist compression? The origin is an interesting and special one: entropy.
That is, the pressure or stiffness of the ideal gas is entirely entropic (rather than enthalpic or energy-based in the case of condensed matter). Nature prefers having the largest number of possible arrangements, which drives the ideal gas to spread to fill a rigid container. There's no energy penalty to this spreading, again in contrast to the case of condensed matter, where a positive surface tension tends to minimize the surface area. Gases don't have a positive surface tension.
In the case of a flexible container, we obtain an equilibrium volume that maximizes the entropy of the universe. Since the gas cools as it inflates its container (because the particles lose energy when they bounce against a retreating boundary), the equilibrium point is reached when the increase in total entropy from increased spacing perfectly balances the decrease in total entropy from cooling.
Alternatively, we can work under an energy framework: the equilibrium volume of the ideal gas is reached when the additional work required to expand the container slightly is in perfect balance with the additional strain energy in the container that such stretching would store.
Does this make sense?
what restricts it from being shrunk? I know it is due to pressure.
Yes. For a flexible container, usually it reaches a volume that allows internal and external pressures to be equal. Otherwise it would change shape.
But I want to know about it in the atomic scale. What stops the molecules from getting much closer?
The molecules are all moving at high speed, and each bounce off the walls imparts a tiny bit of momentum. This sums together to form the pressure on the walls.
The bigger the volume, the farther each particle has to go to hit a wall, and (assuming the same speed), the fewer bounces it can have in a period of time.
The smaller the volume, the faster it can hit the walls.
When you squeeze a balloon, the gas molecules are pushing back on your hand. the smaller the volume you put them in, the faster they can push back.
