What is the celsius temperature of 100.0 g of chlorine gas in a 55.0-l container at 800 mm hg?

To find the Celsius temperature of the chlorine gas under the given conditions, we can use the Ideal Gas Law equation, which is:

PV = nRT

Where: - P is the pressure of the gas in atmosphere (atm) - V is the volume of the gas in liters (L) - n is the number of moles of the gas - R is the ideal gas constant (0.0821 L·atm/mol·K) - T is the temperature of the gas in Kelvin (K)

Firstly, we need to convert the given pressure from mm Hg to atm because the gas constant R is based on atmospheres. 1 atm is equivalent to 760 mm Hg. So:

800 mm Hg * (1 atm / 760 mm Hg) = 1.0526 atm (approximately)

Next, we need to calculate the number of moles of chlorine gas. We know the molar mass of chlorine (Cl_2) is approximately 70.90 g/mol (35.45 g/mol for chlorine atom, since Cl_2 has two atoms we multiply by 2), so we can calculate moles (n) as follows:

n = mass / molar mass n = 100.0 g / 70.90 g/mol n ≈ 1.41 mol (rounded to two decimal places)

Now, we insert our values into the Ideal Gas Law equation to solve for T:

(1.0526 atm)(55.0 L) = (1.41 mol)(0.0821 L·atm/mol·K)T

Solving for T, we get:

T = (1.0526 atm x 55.0 L) / (1.41 mol x 0.0821 L·atm/mol·K) T ≈ (57.893 atm·L) / (0.115791 mol·K) T ≈ 500 K

To convert the temperature from Kelvin to Celsius, we use the formula:

°C = K - 273.15

T(°C) = 500 K - 273.15 T(°C) ≈ 226.85°C

So the temperature of the chlorine gas under the given conditions is approximately 226.85 degrees Celsius