In this problem, we are dealing with one mole of an ideal diatomic gas. We need to determine the final temperature after expansion, given that the process is adiabatic. This means no heat is transferred into or out of the system. Let's solve this step-by-step using thermodynamics principles.
For an adiabatic process, the relation between the initial and final states can be given by:
\(PV^\gamma = \text{constant}\)
where:
The transformation rule for temperature during an adiabatic process for an ideal gas is:
\(\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}\)
where:
Given that the expansion is a doubling of the volume at a constant pressure, we have:
Substituting this into the equations, we have:
\(\frac{T_2}{T_1} = \left(\frac{V_1}{2V_1}\right)^{\gamma-1} = \left(\frac{1}{2}\right)^{0.4}\)
Calculate the above expression:
\(\left(\frac{1}{2}\right)^{0.4} \approx 0.742\)
Thus, we have:
\(T_2 = T_1 \times 0.742\)
Let's assume the initial temperature \(T_1\) to be in Kelvin;
Converting from Celsius (assuming a standard initial temperature, like 27°C which is approximately 300 K for an initial context):
\(300 \times 0.742 = 222.6 \, \text{K}\)
Convert the final temperature to Celsius:
\(222.6 \, \text{K} - 273 \approx -50.4 \,^\circ\text{C}\)
The closest option available is \(-56 \,^\circ\text{C}\), which is the correct answer.
Therefore, upon the adiabatic expansion, the final temperature of the gas is approximately -56°C.

One mole each of \(A_2(g)\) and \(B_2(g)\) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K: \(A_{2}(g)+B_{2}(g) \rightleftharpoons 2AB(g)\). The value of x (missing enthalpy of \(B_2\) or related parameter) is ______ . (Nearest integer)}