To solve this problem, we can use the concept of the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Since the temperature and pressure are unchanged, we can write the equation as:
(P₁)(V₁) = (n₁)(RT) ... Equation 1
where: P₁ = Pressure (unchanged) V₁ = Initial volume = 5.00 L n₁ = Initial number of moles = 0.965 mol
We can rearrange Equation 1 to solve for R, the gas constant:
R = (P₁)(V₁) / (n₁)(T) ... Equation 2
Now, let's find the value of R using the given information and standard conditions. The gas constant, R, is approximately 0.0821 L·atm/(mol·K).
R = (P₁)(V₁) / (n₁)(T) 0.0821 L·atm/(mol·K) = (P₁)(5.00 L) / (0.965 mol)(T)
Since T and P₁ are constant, we can treat their product as a constant and rewrite the equation as:
0.0821 L·atm/(mol·K) = C / (0.965 mol) C = (0.0821 L·atm/(mol·K))(0.965 mol) C ≈ 0.0792 L·atm/K
Now that we have the constant, C, we can use it to find the new volume, V₂, when the number of moles, n₂, is increased to 1.80 mol.
(P₁)(V₂) = (n₂)(RT) V₂ = (n₂)(RT) / P₁ V₂ = (1.80 mol)(0.0792 L·atm/K) / P₁
Since the pressure, P₁, is unchanged and not given in the problem, we can't calculate the exact value of V₂. However, we can determine the ratio of V₂ to V₁:
V₂/V₁ = (1.80 mol)(0.0792 L·atm/K) / (0.965 mol)(0.0792 L·atm/K) V₂/V₁ ≈ 1.87
Therefore, the new volume, V₂, will be approximately 1.87 times the initial volume, V₁:
V₂ ≈ 1.87 * 5.00 L V₂ ≈ 9.35 L
Thus, when the amount of gas is increased to 1.80 mol, the new volume (at an unchanged temperature and pressure) will be approximately 9.35 L.