Question

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? Using the equation P=(nRT)/V

Answer 1

Hopefully this will help you.

Answer 2

Answer:

$P =55.9kPa$

Explanation:

Hello,

In this case, by using the ideal gas law, we obtain the pressure in atm as:

$P=\frac{6.25x10^{-3}mol*(265+273.15)K*0.082\frac{atm*L}{molK} }{0.5000L}=0.552atm$

Thus, in kPa it result:

$P=0.552atm*\frac{101.325kPa}{1atm} =55.9kPa$

Best regards.