Question

A constant-volume gas thermometer is calibrated in dry ice (that is, carbon dioxide in the solid state, which has a temperature of -80.0°C) and in boiling ethyl alcohol (78.0°C). The two pressures are 0.900 atm and 1.635 atm. (a) What Celsius value of absolute zero does the calibration yield? What is the pressure at (b) the freezing point of water and (c) the boiling point of water?

Answer 1

Answer on #69511, Physics / Molecular Physics | Thermodynamics

Question. A constant-volume gas thermometer is calibrated in dry ice (that is, carbon dioxide in the solid state, which has a temperature of $-80.0{}^{\circ}C$) and in boiling ethyl alcohol ($78.0{}^{\circ}C$). The two pressures are 0.900 atm and 1.635 atm. (a) What Celsius value of absolute zero does the calibration yield? What is the pressure at (b) the freezing point of water and (c) the boiling point of water?

Given: $t_1 = -80.0{}^{\circ}C$;

$t_2 = 78.0$;

$p_1 = 0.900\ atm$;

$p_1 = 1.635\ atm$;

$V = const.$

Find: $t_{ab\ zero}$;

$p_f -?$;

$p_b -?$.

Solution. For the ideal gas

$$pV = nRT.$$

Pressure $p$ is directly proportional to $t$: $p = f(t)$, i.e.

$$p = kt + b.$$

Find $k$, and then $b$.

$$0.9 = -80 \cdot k + b;$$ $$1.635 = 78 \cdot k + b$$

Subtract these two equations:

$$0.735 = 158 \cdot k;$$ $$k = 0.0046519\ atm/{}^{\circ}C.$$ $$1.635 = 78 \cdot 0.0046519 + b;$$ $$b = 1.272\ atm.$$

The equation for $p$ and $t$:

$$p = 0.0046519 \cdot t + 1.272.$$

(a) For absolute zero $p = 0$, then

$$0 = 0.0046519 \cdot t_{ab\ zero} + 1.272;$$ $$t_{ab\ zero} = -273.44\ {}^{\circ}C.$$

(b) The freezing point of water is $0\ {}^{\circ}C$, then

$$p_f = 0.0046519 \cdot 0 + 1.272;$$ $$p_f = 1.272\ atm.$$

(c) The boiling point of water is $100\ {}^{\circ}C$, then

$$p_b = 0.0046519 \cdot 100 + 1.272;$$ $$p_b = 1.737\ atm.$$

Answer: (a) $t_{ab\ zero} = -273.44\ {}^{\circ}C$;

(b) $p_f = 1.272\ atm$;

(c) $p_b = 1.737\ atm$.

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