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
V1=300 mLV2=0.360 mLp2=400.0 kPap1V1=p2V2p1=p2V2V1=400.0×0.360300.0=0.48 kPaV_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 \;kPaV1=300mLV2=0.360mLp2=400.0kPap1V1=p2V2p1=V1p2V2=300.0400.0×0.360=0.48kPa
Answer: 0.48 kPa
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