Question

What temperature (in °C) did an ideal gas shift to if it was initially at -10.0 °C at 4.62 atm and 35.0 L and the pressure was changed to 8.71 atm and the volume changed to 15.0 L?

Answer 1

Hi we can use ideal gas equation for this purpose here

PV = nRT, n & R are constant

P1V1 / P2V2 = T1 / T2

P1 = initial pressure = 4.62 atm; V1 = initial volume = 35 L; T1 = -10 ^oC = -10 + 273 = 263 K;

P2 = final pressure = 8.71 atm; V2 = final volume = 15L; T2 = final temp =

T2 = (P2V2 / P1V1) x T1 = [(8.71 atm x 15 L) / (4.62 atm x 35 L)] x 263 K

T1 = 212.498144712 = 212.5 K = 212 - 273 = -60.5 ^oC

Answer:    -60.5 ^oC (negative sign present)

Hope this helped you!

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