In a student experiment, a constant-volume gas thermometer is calibrated in dry
ID: 2240792 • Letter: I
Question
In a student experiment, a constant-volume gas thermometer is calibrated in dry ice (?78.5
In a student experiment, a constant-volume gas thermometer is calibrated in dry ice (?78. 5 degree C) and in boiling allyl alcohol (97 degree C). The separate pressures are 0. 902 atm and 1. 729 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? Use the given calibration data to determine the coefficients in the linear equation that relates pressure and temperature. Then use the pressure for an ideal gas at absolute zero in the linear expression to find the temperature to which it corresponds. degree C What pressure would be found at the freezing point of water? What pressure would be found at the boiling point of water? i dont understand how to do this properlyExplanation / Answer
since there is a linear relation ship between the pressure and temperature, first we calculate the constants A, B by using the given conditions:
1. p1= 0.902atm, T1 = -78.5 c = -78.5 +273 = 194.5k
2. p2 = 1.729 atm, T2 = 97 c. = 97 + 273 = 370k;
relation is P=A + BT by substituting the known values,
0.902= A + B ( 194.5) >>>>>(1);
1.729= A + B (370) >>>>>>>(2);
by solving these two equations ( using calculator), we get, A= -0.014533 and B = 0.00471;
the relation is P = - 0.014533 + (0.00471) * T >>>>>>(3);
NOW WE CAN CALCULATE ANY PRESSURES BY KNOWING THE TEMPERATURE VALUES.
a) absolute zero means T=0;
from eq (3) P = -0.014533 + 0.00471 * 0;
= -0.014533 atm.
b) freezing of water takes place at 0 degree centigrade i.e. at T=273k;
from eq (3) P = -0.014533+ ( 0.00471) * 273;
= 1.2713 atm.
c) boiling of water takes place at 100 degree centigrade i.e. at T= 100+273 = 373k; from eq (3) P= -0.014533 + (0.00471)* 373;
= 1.7423 atm.