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?

Answer: 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