Question

A sample of nitrogen at 20oC was compressed from 300 mL to 0.360 mL and its  new pressure was found to be 400.0 kPa. What was the original pressure in  kPa?

Answer 1

T = 20 ºC = constant

$V_1 = 300 \;mL \\ V_2 = 0.360 \;mL \\ p_2 = 400.0 \;kPa \\ p_1V_1 = p_2V_2 \\ p_1 = \frac{p_2V_2}{V_1} \\ = \frac{400.0 \times 0.360}{300.0} \\ = 0.48 \;kPa$

Answer: 0.48 kPa