Two students in a physics lab have to construct a constant-volume gas thermomete

Question

Two students in a physics lab have to construct a constant-volume gas thermometer as part of their end of year project. They use dry ice (-78.5°C) and boiling ethyl alcohol (78.4°C) to calibrate the thermometer. The corresponding values of pressure are 0.440 atm and 1.575 atm. The students plot a pressure versus temperature graph with pressure on the y-axis and temperature on the x-axis.

(a) Determine the y intercept and slope of the linear graph that relates the pressure to the temperature.

(b) What value of absolute zero do the students get from this calibration?
°C

(c) What pressure readings would this thermometer give for the freezing and boiling points of water?

Explanation / Answer

I will assume that you are assuming the ideal gas law and have not yet been introduced to more complex gas laws. Then we have:

P*V = n*R*T {ideal gas law}

where
P = the absolute pressure of the gas
V = the volume of the gas
T = the absolute temperature of the gas
n = the number of moles of gas
R = the universal gas constant = 8.314472 J/mol/K

The thermometer is constant volume and is (of course) sealed, so that both V and n are constant.
The ideal gas law must apply for the gas at each temperature, so we get:
Pa*V = n*R*Ta
Pb*V = n*R*Tb

However, the temperatures given are in degrees Celsius, not in an absolute temperature scale. We must have that:
Ta = A*ta + B
Tb = A*tb + B
where
ta = -78.5 Celsius
tb = 78.4 Celsius
and A and B are unknown constants that relate the Celsius scale to the absolute scale.

Then we can write the ideal gas law using the Celsius scale as follows:

P*V = n*R*(A*t+B)
P = (n*R*A/V)*t + (n*R*B/V)
P = C*t + D

where we have defined two new unknown constants:
C = n*R*A/V
D = n*R*B/V

Then for the two calibration conditions we have:

Pa = C*ta + D
Pb = C*tb + D

and we now have two equations in the two unknowns C & D, since we know:
ta = -78.5 Celsius
Pa = 0.440 atm
tb = 78.4 Celsius
Pb = 1.575 atm

So we solve for C and D:
1.575 = C*(78.4) + D
0.440 = C*(-78.5) + D

Subtract these two equations to get:
1.575 - 0.440 = C*(78.4+78.5) + (D - D)
1.135 = C*(156.9)
C = 0.007234 atm/Celsius

Use this new C in one of the equations to get D:
1.575 = (0.007234)*(78.4) + D
D = 1.00786 atm

So our equation for P and t becomes:

P = (0.007234)*t + 1.00786

Since you cannot have negative pressure, we must have that at P=0 we have the lowest possible temperature:

0 = (0.007234)*tzero + 1.00786
-1.00786 = 0.007234*tzero
tzero = -139.32 Celsius

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