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?

Which variables are given?

Answer: 55.7 kPa

Explanation:

Given variables are:

Volume of gas (V) = 500.0mL

convert volume in milliliters to liters

(since 1000mL = 1L

500.0mL = 500.0/1000 = 0.5L)

Temperature (T) = 265°C

Convert temperature in celsius to Kelvin

(265°C + 273 = 538K)

Pressure (P) = ?

Number of mole (n) = 6.25 x 10^-3 moles

- Molar gas constant (R) is a constant with a value of 0.0821 atm dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

p x 0.5L = 6.25 x 10^-3 mole 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 = 0.55 atm

The pressure in atmosphere is 0.55. So, convert it to kilopascal (kPa)

Since 1 atm = 101.325 kPa

0.55 atm = Z

cross multiply

Z x 1 atm = 101.325 kPa x 0.55 atm

Z = (101.325 kPa x 0.55 atm) / 1 atm

Z = 55.7 kpa

Thus, the pressure of the gas is 55.7 kilopascals