A sample of gas contains 6.25 × 10-3 mol in a 500.0 mL flask at 265°C. What is the pressure of the gas in kilopascals?

55.9 kilopascals

Explanation:

Given that:

Volume of gas (V) = 500.0mL

[convert 500.0 mL to liters

If 1000 mL = 1 liter

500.0 mL = 500.0/1000 = 0.5 liters]

Temperature of gas (T) = 265°C

Convert 265°C to Kelvin by adding 273

(265°C + 273 = 538K)

Pressure of gas (P) = ?

Number of moles (n) = 6.25 × 10^-3 mol

Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1

To get pressure, apply ideal gas equation

pV = nRT

p x 0.5L = 6.25 × 10^-3 mol x 0.0821 atm L K-1 mol-1 x 538K

0.5L•p = 0.276 atm L

Divide both sides by 0.5L

0.5L•p / 0.5L = 0.276 atm L / 0.5L

p = 0.552 atm

Since pressure is required in kilopascal, convert 0.552 atm to kilopascal

If 1 atm = 101.325 kPa

0..552 atm = Z

cross multiply

Z x 1 atm = 0.552 atm x 101.325 kPa

Z x 1 atm = 55.9 atm•kPa

Z = 55.9 atm•kPa / 1 atm

Z = 55.9 kPa

Thus, the pressure of the gas in kilopascals is 55.9 kPa

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.