Chapter 10, Problem 6P

(a)

To determine

The value of absolute zero in Celsius.

Answer to Problem 6P

The value of absolute zero in Celsius is $-{270}^{\text{ο}}\text{C}$.

Explanation of Solution

Section 1:

Explanation:

To determine: The constants a and b.

Answer:

The constant (a) is $4.7\times {10}^{-3}{\text{atm/}}^{\text{ο}}\text{C}$

The constant (b) is 1.27 atm.

Given Info: The pressure ( $P_1$) is 0.900 atm when temperature ( $T_1$) is $-{78.5}^{\text{ο}}\text{C}$. The pressure ( $P_2$) is 1.635 atm when temperature ( $T_2$) is ${78.0}^{\text{ο}}\text{C}$.

Since, the volume is constant, the P-T graph is a straight line of the form,

$P=aT+b$                            (I)

• P is the pressure.
• T is the temperature.
• a and b are constants.

For the two sets of pressure and temperature,

$P_1=a{T}_1+b$                          (II)

$P_2=a{T}_2+b$                        (III)

From Equations (I) and (II),

$a=\frac{P_1-P_2}{T_1-T_2}$                         (IV)

Substitute 0.900 atm for $P_1$, $-{78.5}^{\text{ο}}\text{C}$ for $T_1$, 1.635 atm for $P_2$ and ${78.0}^{\text{ο}}\text{C}$ for $T_2$ to get a.

$\begin{array}{c}a=\frac{\left(0.900\text{atm}\right)-\left(1.635\text{atm}\right)}{\left[-{78.5}^{\text{ο}}\text{C}\right]-\left[{78.0}^{\text{ο}}\text{C}\right]}\\ =4.7\times {10}^{-3}{\text{atm/}}^{\text{ο}}\text{C}\end{array}$

From Equation (II),

$b=P_1-a{T}_1$

Substitute 0.900 atm for $P_1$, $-{78.5}^{\text{ο}}\text{C}$ for $T_1$ and $4.7\times {10}^{-3}{\text{atm/}}^{\text{ο}}\text{C}$ for a.

$\begin{array}{c}b=\left(0.900\text{atm}\right)-\left[4.7\times {10}^{-3}{\text{atm/}}^{\text{ο}}\text{C}\right]\left[-{78.5}^{\text{ο}}\text{C}\right]\\ =1.27\text{atm}\end{array}$

The constant (a) is $4.7\times {10}^{-3}{\text{atm/}}^{\text{ο}}\text{C}$

The constant (b) is 1.27 atm.

Conclusion:

From Equation (I), the temperature is,

$T=\frac{b-P}{a}$

At absolute zero, pressure is zero.

Substitute 0 atm for P, $4.7\times {10}^{-3}{\text{atm/}}^{\text{ο}}\text{C}$ for a and 1.27 atm for b to get T.

$\begin{array}{c}T=\frac{\left(0.\text{atm}\right)-\left(1.27\text{atm}\right)}{4.7\times {10}^{-3}{\text{atm/}}^{\text{ο}}\text{C}}\\ =-{270.21}^{\text{ο}}\text{C}\end{array}$

On Rounding off,

$T=-{270}^{\text{ο}}\text{C}$

(b)

To determine

The pressure.

Answer to Problem 6P

The pressure is 1.27 atm.

Explanation of Solution

From Equation (I) of (a),

$P=aT+b$

At the freezing point, temperature is zero.

Substitute $0^{\text{ο}}\text{C}$ for T, $4.7\times {10}^{-3}{\text{atm/}}^{\text{ο}}\text{C}$ for a and 1.27 atm for b to get P.

$\begin{array}{c}P=\left(4.7\times {10}^{-3}{\text{atm/}}^{\text{ο}}\text{C}\right)\left(0^{\text{ο}}\text{C}\right)+\left(1.27\text{atm}\right)\\ =1.27\text{atm}\end{array}$

Conclusion:

The pressure is 1.27 atm.

(c)

To determine

The pressure.

Answer to Problem 6P

The pressure is 1.74 atm.

Explanation of Solution

From Equation (I) of (a),

$P=aT+b$

At the boiling point, temperature is ${100}^{\text{ο}}\text{C}$.

Substitute ${100}^{\text{ο}}\text{C}$ for T, $4.7\times {10}^{-3}{\text{atm/}}^{\text{ο}}\text{C}$ for a and 1.27 atm for b to get P.

$\begin{array}{c}P=\left(4.7\times {10}^{-3}{\text{atm/}}^{\text{ο}}\text{C}\right)\left({100}^{\text{ο}}\text{C}\right)+\left(1.27\text{atm}\right)\\ =1.74\text{atm}\end{array}$

Conclusion:

The pressure is 1.74 atm.