Does this mean above and below the coexistence curve of liquid and gas, there is only pure liquid or pure gas phase present?
Yes. For example, water in vapor-liquid equilibrium at $\ce{1 atm}$ can only exist at $\ce{100 °C}$. If you change the $P$ while keeping $T$, or you change $T$ while keeping $P$, you will have water in the gas or liquid state.
In another case, if we have a closed container with a piston to ensure a constant external 1 atm pressure on the container filled with water at 20°C, will this mean there will be no vapour pressure/evaporation in the system, as the pressure and temperature conditions correspond to pure liquid on the phase diagram of water...
Yes. If you keep water at $\ce{1 atm}$ and $\ce{20 °C}$, as the phase diagram points out, water can only exist as a liquid.
Further, if a liquid in closed container is in equilibrium with its vapour (rate of condensation = rate of evaporation), is it true that the point corresponding to its (P, T) will necessarily lie on the coexistence curve (this does seem true by definition, but I might be wrong).
Yes. Phase transitions for pure fluids take place at constant pressure and temperature. For example, if you heat saturated liquid at $T = 100 °C$, saturated gas will form in presence of (now in a less quantity) saturated liquid. If more heat is added, then you will end up only with saturated gas. All this changes take place at constant temperature. However, heat a bit more, and you will have a superheated gas but now $T \neq 100 °C$.
The process must be done sufficiently slow so that thermodynamics still apply and our statements hold. Also, this evolution cannot be traced in the P-T diagram, but you can clearly describe it in the P-V diagram.
And can we say that a liquid will evaporate if and only if the conditions of pressure and temperature lie on the coexistence curve, and otherwise it will exist either as a pure liquid or as a pure gas? If yes, isn't it a contradiction to evaporation under standard conditions?
Yes. Remember that a pure fluid at VLE has $1$ degree of freedom. So, as long as you can mantain the pressure, more precisely the saturation pressure, it will evaporate at constant $P$ and $T$. If the pressure is changed, you move out of the V-L coexistence curve and another process will take place.
Some Mathematics of Evaporation
I see that you are interested in the process of evaporation. Thermodynamics will tell you nothing about rates, but we can make up a simple situation that you may like. Lets consider the following figure:
Say we heat an initial number of moles $n_{L0}$ of saturated liquid from the bottom at a heat rate $\dot{Q}$. As a consequence, we have an intlet molar flow $\dot{n}$ from the liquid phase to the gas phase. As the liquid phase evaporates, the energy that is losing has an enthalpy of $h_L^\pu{vap}$. We will apply the laws for both separate systems, and thus, we have to apply the laws for open systems. We disregard energy such as potential energy and no work is being done.
The mole balances for both phases yield
$$
\frac{dn_L}{dt} = -\dot{n} \hspace{2 cm}
\frac{dn_G}{dt} = \dot{n} \tag{1,2}
$$
The 1st law yields
$$
\frac{d(n_Lu_L)}{dt} = \dot{Q} -\dot{n}h_L^\pu{vap} \hspace{2 cm}
\frac{d(n_Gu_G)}{dt} = \dot{n}h_L^\pu{vap} \tag{3,4}
$$
Principal hypothesis (the one that interests you): the pressure of both systems (and thus the temperature by Gibbs phase rule) is mantained. This is, the chemical potential of both phases are equal. Therefore, all enthalpies and internal energies are constant and may be taken out from the time derivatives. With Eq. (4) we can say the following
\begin{align}
u_G^\pu{vap}\frac{dn_G}{dt} &= \dot{n}h_L^\pu{vap} \hspace{1 cm} \text{(use Eq. (2))} \\
u_G^\pu{vap}(\dot{n}) &= \dot{n}h_L^\pu{vap} \rightarrow
h_L^\pu{vap} = u_G^\pu{vap} \tag{5}
\end{align}
And Eq. (3) tells us the following
\begin{align}
u_L^\pu{vap}\frac{dn_L}{dt} &= \dot{Q} -\dot{n}h_L^\pu{vap} \hspace{1 cm} \text{(use Eqs. (1) and (5))} \\
u_L^\pu{vap}(-\dot{n}) &= \dot{Q} -\dot{n}u_G^\pu{vap} \\
\dot{Q} &= \dot{n}(u_G^\pu{vap} - u_L^\pu{vap}) \rightarrow
\boxed{\dot{n} = \frac{\dot{Q}}{\Delta_\pu{vap} u}} \tag{6}
\end{align}
Eq.(6) may seem kind of obvious, but we have to prove things. Consequently, if the heat rate $\dot{Q}$ is constant, so will be the molar flow rate $\dot{n}$.
Of interest in thermodynamics is the quality, i.e., the mole fraction of gas in our system. Eqs. (1) and (2) integrated yield
$$ n_L(t) = n_{L0} - \dot{n}t \hspace{2 cm} n_G(t) = \dot{n}t \tag{7,8} $$
Eq. (7) states that we will run out of liquid when $t = n_{L0}/\dot{n}$. Finally, combining Eqs. (6-8)
\begin{align}
x(t) &= \frac{n_G(t)}{n_L(t) + n_G(t)} \\
x(t) &= \frac{\dot{n}t}{n_{L0}} \rightarrow
\boxed{x(t) = \bigg(\frac{\dot{Q}}{n_{L0}\Delta_\pu{vap}u}\bigg)t \hspace{0.5 cm}
0 \leq t \leq \frac{\Delta_\pu{vap}u}{n_{L0}\dot{Q}}} \tag{9}
\end{align}
Eq. (9) is the most simple but also beautiful result that we can get from an evaporation process if thermodynamics holds. Everything of it makes sense:
- The quality will be an increasing linear function.
- Higher the internal energy of evaporation, the harder it will be to evaporate the liquid. We may regard it, with a cautious distance from kinetic processes, as a kind of "thermal resistance".
- Higher the heat rate, more rapidly the quality will increase.
