Chapter 10, Problem 60AP

A 20.0-L tank of carbon dioxide gas (CO₂) is at a pressure of 9.50 × 10⁵ Pa and temperature of 19.0°C (a) Calculate the temperature of the gas in Kelvin. (b) Use the ideal gas law to calculate the number of moles of gas in the tank. (c) Use the periodic table to compute the molecular weight of carbon dioxide, expressing it in grams per mole. (d) Obtain the number of grains of carbon dioxide in the tank. (e) A fire breaks out, raising the ambient temperature by 224.0 K while 82.0 g of gas leak out of the tank. Calculate the new temperature and the number of moles of gas remaining in the tank. (f) Using a technique analogous to that in Example 10.6b, find a symbolic expression for the final pressure, neglecting the change in volume of the tank. (g) Calculate the final pressure in the tank as a result of the fire and leakage.

To determine

The temperature in Kelvin scale.

Answer to Problem 60AP

The temperature in Kelvin scale is 292 K.

Explanation of Solution

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is ${19.0}^{\text{ο}}\text{C}$.

The pressure (P) is $9.50\times {10}^5\text{Pa}$

Formula to calculate the temperature in Kelvin scale is,

$T_K=\left(T+273.15\right)\text{K}$

Substitute ${19.0}^{\text{ο}}\text{C}$ for T in the above equation.

$\begin{array}{c}{T}_K=\left(19.0+273.15\right)\text{K}\\ =\text{292}\text{K}\end{array}$

Conclusion:

The temperature in Kelvin scale is 292 K.

To determine

The number of moles of gas.

Answer to Problem 60AP

The number of moles of gas is 7.83 mol.

Explanation of Solution

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is ${19.0}^{\text{ο}}\text{C}$.

The pressure (P) is $9.50\times {10}^5\text{Pa}$

From Ideal gas equation, formula to calculate the number of moles of gas is,

$n=\frac{PV}{RT}$

• R is the gas constant.

Substitute 292 K for T, $9.50\times {10}^5\text{Pa}$ for P, 20.0 L for V and $8.31{\text{Jmol}}^{-1}{\text{K}}^{-1}$ for R in the above equation.

$\begin{array}{c}n=\frac{\left(9.50\times {10}^5\text{Pa}\right)\left(20.0\text{L}\right)}{\left(8.31{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\left(292\text{K}\right)}\\ =\frac{\left(9.50\times {10}^5\text{Pa}\right)\left(20.0\times {10}^{-3}{\text{m}}^{-3}\right)}{\left(8.31{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\left(292\text{K}\right)}\\ =7.83\text{mol}\end{array}$

Conclusion:

The number of moles of gas is 7.83 mol.

To determine

The molecular weight of carbon dioxide (M).

Answer to Problem 60AP

The molecular weight of carbon dioxide is 44.0 g/mol.

Explanation of Solution

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is ${19.0}^{\text{ο}}\text{C}$.

The pressure (P) is $9.50\times {10}^5\text{Pa}$

The formula of carbon dioxide is ${\text{CO}}_{\text{2}}$. Therefore, the molecular weight is equal to the molecular weight of carbon and two oxygen atoms.

The molecular weight of carbon is 12.0 g/mol. The molecular weight of oxygen is 16.0 g/mol.

The molecular weight of ${\text{CO}}_{\text{2}}$ is,

$\begin{array}{c}M=\left(12.0\text{g/mol}\right)+2\left(16.0\text{g/mol}\right)\\ =44.0\text{g/mol}\end{array}$

Conclusion:

The molecular weight of carbon dioxide is 44.0 g/mol.

To determine

The number of grams of carbon dioxide (m).

Answer to Problem 60AP

The number of grams of carbon dioxide is 345 g.

Explanation of Solution

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is ${19.0}^{\text{ο}}\text{C}$.

The pressure (P) is $9.50\times {10}^5\text{Pa}$

From (b), $n=7.93\text{mol}$

From (c), $M=44.0\text{g/mol}$.

Formula to calculate the number of grams of carbon dioxide is,

$m=nM$

Substitute 7.93 mol for n and 44.0 g/mol for M in the above equation to get m.

$\begin{array}{c}m=\left(7.93\text{mol}\right)\left(44.0\text{g/mol}\right)\\ =345\text{g}\end{array}$

Conclusion:

The number of grams of carbon dioxide is 345 g.

To determine

The new temperature and number of moles of gas after the leakage.

Answer to Problem 60AP

The new temperature is 516 K.

The number of moles of gas after the leakage is 5.98 mol.

Explanation of Solution

Section 1:

To determine: The new temperature.

Answer: The new temperature is 516 K.

Explanation:

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is ${19.0}^{\text{ο}}\text{C}$.

The pressure (P) is $9.50\times {10}^5\text{Pa}$

The temperature raises by 224.0 K. 82.0 g of gas leaks out.

From (a), $T=292\text{K}$.

Formula to calculate the new temperature is,

$T_N=T+\left(224.0\text{K}\right)$

Substitute 292 K for T in the above equation to get m.

$\begin{array}{l}{T}_N=\left(292\text{K}\right)+\left(224.0\text{K}\right)\\ =516\text{K}\end{array}$

The new temperature is 516 K.

Section 2:

To determine: The number of moles of gas after the leakage.

Answer: The number of moles of gas after the leakage is 5.98 mol.

Explanation:

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is ${19.0}^{\text{ο}}\text{C}$.

The pressure (P) is $9.50\times {10}^5\text{Pa}$

The temperature raises by 224.0 K. 82.0 g of gas leaks out.

From (d), $m=345\text{g}$

From (c), $M=44.0\text{g/mol}$.

Formula to calculate the number of moles of gas after the leakage is,

$n^{\text{'}}=\frac{m-\left(82.0\text{g}\right)}{M}$

Substitute 345 g for m and 44.0 g/mol for M in the above equation.

$\begin{array}{l}{n}^{\text{'}}=\frac{\left(345\text{g}\right)-\left(82.0\text{g}\right)}{\text{44}\text{.0}\text{g/mol}}\\ =5.98\text{mol}\end{array}$

The number of moles of gas after the leakage is 5.98 mol.

Conclusion:

The new temperature is 516 K.

The number of moles of gas after the leakage is 5.98 mol.

To determine

The expression for final pressure.

Answer to Problem 60AP

The expression for final pressure is $P_i\left(\frac{V_i}{V_f}\right)\left(\frac{n_f}{n_i}\right)\left(\frac{T_f}{T_i}\right)$.

Explanation of Solution

Given info:

From the ideal gas equation,

$PV=nRT$

Therefore,

$\frac{P_f{V}_f}{P_i{V}_i}=\frac{n_{f}R{T}_f}{n_{i}R{T}_i}$

• $P_i$ and $P_f$ are the initial and final pressure.
• $V_i$ and $V_f$ are the initial and final volume.
• $n_i$ and $n_f$ are the initial and final number of moles of gas.
• $T_i$ and $T_f$ are the initial and final temperature.

On Re-arranging the above equation,

$P_f=P_i\left(\frac{V_i}{V_f}\right)\left(\frac{n_f}{n_i}\right)\left(\frac{T_f}{T_i}\right)$

Conclusion:

The expression for final pressure is $P_i\left(\frac{V_i}{V_f}\right)\left(\frac{n_f}{n_i}\right)\left(\frac{T_f}{T_i}\right)$.

To determine

The final pressure in the tank as the result of fire and leakage.

Answer to Problem 60AP

The final pressure in the tank as the result of fire and leakage is $1.28\times {10}^6\text{Pa}$

Explanation of Solution

Given info:

Volume of the tank (V) is 20.0 L.

The temperature (T) is ${19.0}^{\text{ο}}\text{C}$.

The pressure (P) is $9.50\times {10}^5\text{Pa}$

The temperature rises by 224.0 K. 82.0 g of gas leaks out.

From (f),

$P_f=P_i\left(\frac{V_i}{V_f}\right)\left(\frac{n_f}{n_i}\right)\left(\frac{T_f}{T_i}\right)$ (II)

Since the volume is the same, $V_i=V_f$.

Substitute 5.98 mol for $n_f$, 7.93 mol for $n_i$, 516 K for $T_f$, 292 K for $T_i$, $9.50\times {10}^5\text{Pa}$ for $P_i$ and $V_i$ for $V_f$ in the Equation (II) to get $P_f$.

$\begin{array}{c}{P}_f={\left(9.50\times {10}^5\text{Pa}\right)}_i\left(\frac{V_i}{V_i}\right)\left(\frac{5.98\text{mol}}{7.93\text{mol}}\right)\left(\frac{516\text{K}}{292\text{K}}\right)\\ =1.28\times {10}^6\text{Pa}\end{array}$

Conclusion:

The final pressure in the tank as the result of fire and leakage is $1.28\times {10}^6\text{Pa}$.