A constant-volume gas thermometer is calibrated in dry ice (-80 C) and in boilin

Question

A constant-volume gas thermometer is calibrated in dry ice (-80 C) and in boiling ethyl alcohol (78 C). The respective pressures are 0.900 atm and 1.635 atm. What value of absolute zero does the calibration yield; and what pressures are at the freezing and boiling points of water?

Explanation / Answer

We know that P = m T + Po Where m = slope Slope m = (1.635 - 0.900) / (78 - (-80)) = 0.004651 Therefore p = 0.004651 T + po ------------------------------------------------------------------------------------------------------- What value of absolute zero does the calibration yield 1.635 = 0.004651 * 78 + po ==> po = 1.272 atm So that p = 0.004651 T + 1.272 We need to find the temp when P = 0 So we have 0 = 0.004651T + 1.272 ==> T = -273.46 oC ----------------------------------------------------------------------------------------------------- what pressures are at the freezing and boiling points of water? At freezing point T = 0 oc ==> P = 1.272 atm At boiling point T = 100 oC ==> P = 0.004651 * 100 + 1.272 = 1.7371 atm

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