Answer & Explanation:
We can use the combined gas law to solve this problem:
(P₁V₁)/T₁ = (P₂V₂)/T₂
where P is pressure, V is volume, and T is temperature in Kelvin.
First, let's convert the temperatures to Kelvin:
T₁ = 25.0 °C + 273.15 = 298.15 K
T₂ = 10.0 °C + 273.15 = 283.15 K
Next, plug in the values we know:
(P₁)(752 mL)/(298.15 K) = (P₂)(V₂)/(283.15 K)
Since the pressure is constant, we can simplify the equation:
(P₁)(752 mL)/(298.15 K) = (P₂)(V₂)/(283.15 K)
(P₁)(752 mL)(283.15 K) = (P₂)(298.15 K)(V₂)
(P₁)(752 mL)(283.15 K)/(298.15 K) = P₂V₂
V₂ = (P₁)(752 mL)(283.15 K)/(P₂)(298.15 K)
We don't know the pressure, so we can't solve for V₂ directly. However, if we assume that the pressure stays the same, we can use the ideal gas law to find the pressure:
PV = nRT
where n is the number of moles of gas and R is the gas constant.
We know that neon is a monatomic gas with a molar mass of 20.18 g/mol. Let's assume we have one mole of neon gas:
PV = (1 mol)(8.314 J/(mol·K))(283.15 K)
P = (8.314 J/(mol·K))(283.15 K)/V
P = 2355 Pa
Now we can solve for V₂:
V₂ = (P₁)(752 mL)(283.15 K)/(P₂)(298.15 K)
V₂ = (1 atm)(752 mL)(283.15 K)/(2355 Pa)(298.15 K)
V₂ = 0.822 L or 822 mL (rounded to three significant figures)
Therefore, the volume of the neon gas at 10.0 °C and constant pressure should be approximately 822 mL.