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?
T = 20 ºC = constant
"V_1 = 300 \\;mL \\\\\n\nV_2 = 0.360 \\;mL \\\\\n\np_2 = 400.0 \\;kPa \\\\\n\np_1V_1 = p_2V_2 \\\\\n\np_1 = \\frac{p_2V_2}{V_1} \\\\\n\n= \\frac{400.0 \\times 0.360}{300.0} \\\\\n\n= 0.48 \\;kPa"
Answer: 0.48 kPa
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