Answer:
1) The absolute zero temperature is -272.74 °C
2) The absolute zero temperature is -269.91°C
Explanation:
1) The specific volume of air at the given temperatures are;
At -85°C, v = 1/(1.877) = 0.533 dm³/g, the pressure =
At 0°C, v = 1/1.294 ≈ 0.773 dm³/g
At 100°C, v = 1/0.946 ≈ 1.057 dm³/g
We therefore have;
v₁ = 0.533 T₁ = 188.15 K
v₂ = 0.773 T₂ = 273.15 K
v₃ = 1.057 T₃ = 373.15 K
The slope of the graph formed by the above data is therefore given as follows;
m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K
The equation is therefor;
v - 0.533 = 0.00284×(T - 188.15)
v = 0.00284×T - 0.534 + 0.533 = 0.00284×T - 0.001167
v = 0.00284×T - 0.001167
Therefore, when the temperature, at absolute 0, we have;
v = 0
Which gives;
0 = 0.00284×T - 0.001167
0.00284×T = 0.001167
T = 0.001167/0.00284 ≈ 0.411 K
Which is 0.411 -273.15 ≈ -272.74 °C
The absolute zero temperature is -272.74 °C
2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm
The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)
The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;
0°C = 0 + 273.15 K = 273.15 K
Therefore, the equation of the graph can be presented as follows;
v - 20.00 = 0.0741 × (T - 273.15)
Which gives;
v = 0.0741·T - 0.0741 ×(273.15) + 20
v = 0.0741·T - 20.2404125 + 20
v = 0.0741·T - 0.240415
Therefore at absolute 0, v = 0, we have;
0 = 0.0741·T - 0.240415
0.0741·T = 0.240415
T = 0.240415/0.0741 = 3.2445 K
The temperature in degrees is therefore;
3.2445 K - 273.15 ≈ -269.91°C
The absolute zero temperature is therefore, -269.91°C
