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Answer :
The temperature is calibrated using dry ice and ethanol and the temperatures should actually be listed as:
dry ice, T = -78.5 °C, P = 0.900 atm
ethanol, T = 78 °C, P = 1.635 atm
We can use the formula P = A + BT where A and B are constants we can solve for:
0.900 atm = A + B(-78.5)
A = 0.900 + 78.5B
1.635 = A + B(78)
A = 1.635 - 78B
1.635 - 78B = 0.900 + 78.5B
156.5B = 0.735
B = 0.004696
A = 1.635 - 78(0.004696)
A = 1.269
We have values for A and B, and we also know that at absolute zero, the pressure is equal to zero. Therefore, we can solve for the temperature at absolute zero using the calibrated values of A and B:
P = A + BT
0 = 1.269 + (0.004696)T
0.004696T = -1.269
T = (-1.269/0.004696)
T = -270.2 °C
The calibrated temperature provides a value of absolute zero as -270.2 °C when we know the true value is -273.15 °C.
dry ice, T = -78.5 °C, P = 0.900 atm
ethanol, T = 78 °C, P = 1.635 atm
We can use the formula P = A + BT where A and B are constants we can solve for:
0.900 atm = A + B(-78.5)
A = 0.900 + 78.5B
1.635 = A + B(78)
A = 1.635 - 78B
1.635 - 78B = 0.900 + 78.5B
156.5B = 0.735
B = 0.004696
A = 1.635 - 78(0.004696)
A = 1.269
We have values for A and B, and we also know that at absolute zero, the pressure is equal to zero. Therefore, we can solve for the temperature at absolute zero using the calibrated values of A and B:
P = A + BT
0 = 1.269 + (0.004696)T
0.004696T = -1.269
T = (-1.269/0.004696)
T = -270.2 °C
The calibrated temperature provides a value of absolute zero as -270.2 °C when we know the true value is -273.15 °C.
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