Question

# In a student experiment, a constant-volume gas thermometer is calibrated in dry ice (?78.5°C) and in...

In a student experiment, a constant-volume gas thermometer is calibrated in dry ice (?78.5°C) and in boiling pentane (36.1°C). The separate pressures are 0.902 atm and 1.442 atm. Hint: Use the linear relationship P = A + BT, where A and B are constants.

(a) What value of absolute zero does the calibration yield?
°C

(b) What pressure would be found at the freezing point of water?
atm

(c) What pressure would be found at the boiling point of water?
atm

0.902 atm = A + B(-78.5 + 273K) = A + B*194.5K

1.442 atm = A + B(36.1 + 273K) = A + B*309.1K

Solving for simultaneous equations yields

A = -0.0144937 atm

B = 0.00471205 atm/°K

Therefore

P = -0.0144937 atm + (0.00471205 atm/K)*T

(a) At absolute zero, there is zero pressure (P = 0), therefore T = (0. 0144937 atm)/( 0.00471205 atm/K) = 3.08K or -270C

(b) P = -0.0144937 atm + (0.00471205 atm/K)*(0 + 273°K) = 1.272 atm

(c) P =-0.0144937 atm + (0.00471205 atm/K)*(100 + 273°K) = 1.743 atm