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

Question

Rocket

Chemistry

31 December, 22:32

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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Answers (1)

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Kayla · Answer 1

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)