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  • 21-02-2020
  • Chemistry
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A sample of Kr at 190.21 °C experiences a change in volume from 36.367 L to 89.148 L. If its new pressure is 5.4862 atm at -88.82 °C, what was its original pressure in torr

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Abu99
Abu99 Abu99
  • 21-02-2020

Answer:

10220.9T (to 6 sig' fig's)

Explanation:

PV = nRT ← Ideal Gas Equation

(5.4862)(89.148) = n(8.31)(190.21 + 273)

489.08.... = 3849.2...(n)

n = 0.127....

P(36.367) = (0.127....)(8.31)(463.21)

36.367(P) = 489.08....

P = 13.448.... atm's

And since 1 atm = 760 Torr,

P = 13.448... × 760

P = 10220.9051 T (→ 10220.9 T)

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