Answer:
The volume of the gas is approximately 4.41 liters
Explanation:
The details of the data of the Neon gas are;
The number of moles of Neon gas present, n = 11.6 moles
The temperature of the sample of Neon gas, T = 120 K
The pressure of the sample of the Neon gas, P = 25.6 atm
By the ideal gas equation, we have;
P·V = n·R·T
Where;
R = The universal gal constant = 0.08205 L·atm·mol⁻¹·K⁻¹
Therefore, we get;
V = n·R·T/P
Which gives;
V = 11.6 moles × 0.08205 L·atm·mol⁻¹·K⁻¹ × 120 K/(25.9 atm) ≈ 4.4097915 L
The volume of the gas, V ≈ 4.41 L.
