The CO2 dissolved in the drink is at equilibrium with the CO2 in the 5 atm pure CO2 gas phase. So the chemical potential of the CO2 dissolved in the drink must have the same chemical potential (partial molar Gibbs free energy) as the CO2 gas.

Background

Mixtures

The chemical potential of a component in a mixture $mu_j$ is obtained from its pure value $mu_j^o$ and temperature $T$ as

$$ mu_j = mu_j^o + RTln(a_j) $$

where $a_j$ is the component activity. The typical forms for gases (with pressure $p$) or liquids (with molar concentrations $C$) are

$$a_j = phi_j p_j / p_j^o mathrm{or} a_j = gamma_j C_j/C_j^o$$

where $phi_j$ is a fugacity coefficient and $gamma_j$ is a (molar based) activity coefficient. The reference values $p_j^o, C_j^o$ are unity with the units of pressure or molarity.

Pure components have $phi_j = 1, gamma_j = 1$. Ideal solutions (e.g. ideal gases) also have the same values.

Phase Equilibria

The criteria for chemical equilibrium of a mixture in multiple phases is that $mu_j$ is the same in all phases. The concentrations of species in different phases can often be related to each other through specific relationships. Examples for vapor-liquid systems include Raoult's law or Henry's law.

Specifics

For what follows, I will use D to designate carbon dioxide.

Part 1

The first part has a gas dissolved in a liquid. The solution is a mixture. The chemical potential of the gas component follows from above for mixtures. We have no idea whether the solution is ideal or not. But, we know that the gas in the liquid will be in chemical equilibrium with a pure gas at a certain pressure. This allows us to write the expression for the chemical potential of the gas in the liquid.

$$ mu_{D,liq} = mu_{D,gas}(p = 5 mathrm{atm})$$

Assume that pure D is ideal even at the higher pressure to obtain

$$mu_{D,gas}(p = 5 mathrm{atm}) = mu_{D,gas}(p = 1 mathrm{atm}) + RTln(p(5~mathrm{atm}))$$

The first question asks for the chemical potential of D in the liquid relative to gas at 1 atm.

Part 2

We open the cap at release the 5 atm inside the bottle lid. The gas does not leave the liquid. This becomes a meta-stable (non-equilibrium) system. The chemical potential of the gas in the liquid does not change since none of it leaves the liquid.

Part 3

We put the cap on and allow equilibrium to be established. Gas D leaves the liquid and enters to the gas phase above the liquid to a volume $V_e$. This is an initial state to final state problem. In the initial state, we have a gas at $p_o = 1$ atm with no D above a liquid with a concentration of D defined by the Henry's law $C_{D,o} = Hp_D$ for $p_D = 5$ atm. At the final state, we have a gas at a total final pressure $p_{T,f} = p_o + Delta p$ and a liquid with $C_{D,f} = H p_{D,f}$ where the partial pressure of D in the gas phase determined by the total pressure and mole fraction. We make a link between the pressure change and the concentration change using a mole balance and the ideal gas law

$$Delta n_{D,gas} = Delta p V_e/RT = Delta n_{D,liq} = (C_{D,f} - C_{Do})V_{liq}$$

The new chemical potential relative to pure gas at 1 atm $mu_{D,gas} - mu_{D,1~mathrm{atm}}$ is related to partial pressure $p_{D,f}$ at equilibrium. The numerical value can be found by algebraic manipulation on the above expressions with substitutions from the values given in the problem and an assumption of room temperature.

We have all that we need to solve the third part.

Iterations

Assume that, each time the bottle cap is opened, the carbon dioxide is removed from the gas phase and we start again on step 3 above. In theory, we would need an infinite number of these operations to completely remove all carbon dioxide from the liquid. In practice, going flat is the point where the concentration of D in the liquid can no longer sustain it being able to leave to a gas that is at 1 atm.