How many moles of Br2 are present in 500mL of vapor when Br2(l) and Br2(g) are in equilibrium at 33 degrees celcius? (Pvap Br2 at 33C = 299 mm Hg)

n = PV/RT

P in atm
V in L
T in kelvin.

To determine the number of moles of Br2 present in 500 mL of vapor, we need to use the ideal gas law equation:

PV = nRT

Where:
P is the pressure of the gas (299 mm Hg)
V is the volume of the gas (convert 500 mL to liters by dividing by 1000: 0.5 L)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin (33°C + 273.15 = 306.15 K)

Rearranging the equation, we have:

n = PV / RT

Now, let's plug in the values:

n = (299 mm Hg * 0.5 L) / (0.0821 L·atm/mol·K * 306.15 K)
n = (149.5 mm Hg * L) / (25.24 L·atm/mol·K)

To convert mm Hg to atm, divide by 760:

n = (149.5 mm Hg * L) / (25.24 * (760 mm Hg/atm) * 1 mol)
n = 0.03 mol

Therefore, there are approximately 0.03 moles of Br2 present in 500 mL of vapor at equilibrium.