Answer to Question #155025 in General Chemistry for john

Question #155025

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?

1
Expert's answer
2021-01-12T07:04:09-0500

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 \;kPa

Answer: 0.48 kPa


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