Chapter 10, Problem 44P

A sealed cubical container 20.0 cm on a side contains a gas with three times Avogadro’s number of neon atoms at a temperature of 20.0°C. (a) Find the internal energy of the gas. (b) Find the total translational kinetic energy of the gas. (c) Calculate the average kinetic energy per atom, (d) Use Equation 10.13 to calculate the gas pressure. (e) Calculate the gas pressure using the ideal gas law (Eq. 10.8).

To determine

The internal energy of the gas.

Answer to Problem 44P

The internal energy of the gas is $1.1\times {10}^4\text{J}$.

Explanation of Solution

Given info: The number of moles (n) is 3. The temperature (T) is ${20}^{\text{ο}}\text{C}$. The side of the cube(r) is 20 cm. Molar mass of neon gas is $20.180\times {10}^{-3}\text{kg/mol}$.

Formula to calculate the internal energy of the gas is,

$U=\frac{3}{2}nRT$

• R is the gas constant.

Substitute 3 mol for n, ${20}^{\text{ο}}\text{C}$ for T and $8.31{\text{Jmol}}^{-1}{\text{K}}^{-1}$ for R in the above equation to get U.

$\begin{array}{c}U=\frac{3}{2}\left(3\text{mol}\right)\left(8.31{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\left({20}^{\text{ο}}\text{C}\right)\\ =\frac{3}{2}\left(3\text{mol}\right)\left(8.31{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\left(20\text{+273}\text{K}\right)\\ =1.1\times {10}^4\text{J}\end{array}$

Conclusion:

The internal energy of the gas is $1.1\times {10}^4\text{J}$.

To determine

The total translational kinetic energy of the gas ( ${\left(KE\right)}_{total}$).

Answer to Problem 44P

The total translational kinetic energy of the gas is $1.1\times {10}^4\text{J}$.

Explanation of Solution

Given info: The number of moles (n) is 3. The temperature (T) is ${20}^{\text{ο}}\text{C}$. The side of the cube(r) is 20 cm. Molar mass of neon gas is $20.180\times {10}^{-3}\text{kg/mol}$.

The total translational kinetic energy is equal to the internal energy. Therefore,

${\left(KE\right)}_{total}=U$

From (a), $U=1.1\times {10}^4\text{J}$.

Conclusion:

The internal energy of the gas is $1.1\times {10}^4\text{J}$.

To determine

The average kinetic energy per atom ( ${\left(KE\right)}_{avg}$).

Answer to Problem 44P

The average kinetic energy per atom is $6.07\times {10}^{-21}\text{J}$.

Explanation of Solution

Given info: The number of moles (n) is 3. The temperature (T) is ${20}^{\text{ο}}\text{C}$. The side of the cube(r) is 20 cm. Molar mass of neon gas is $20.180\times {10}^{-3}\text{kg/mol}$.

Formula to calculate the average kinetic energy per atom is,

${\left(KE\right)}_{avg}=\frac{{\left(KE\right)}_{total}}{n{N}_A}$

• $N_A$ is the Avogadro number.

Substitute $1.1\times {10}^4\text{J}$ for ${\left(KE\right)}_{total}$, 3 mol for n and $6.02\times {10}^{23}\text{atoms/mol}$ for $N_A$ to get ${\left(KE\right)}_{avg}$.

$\begin{array}{c}{\left(KE\right)}_{avg}=\frac{1.1\times {10}^4\text{J}}{\left(3\text{mol}\right)\left(6.02\times {10}^{23}\text{atoms/mol}\right)}\\ =6.07\times {10}^{-21}\text{J}\end{array}$

Conclusion:

The average kinetic energy per atom is $6.07\times {10}^{-21}\text{J}$.

To determine

The gas pressure.

Answer to Problem 44P

The gas pressure is $9.14\times {10}^5\text{Pa}$.

Explanation of Solution

Given info: The number of moles (n) is 3. The temperature (T) is ${20}^{\text{ο}}\text{C}$. The side of the cube(r) is 20 cm. Molar mass of neon gas is $20.180\times {10}^{-3}\text{kg/mol}$.

Formula to calculate the gas pressure is,

$P=\frac{2}{3}\left(\frac{n{N}_A}{V}\right){\left(KE\right)}_{avg}$ (I)

• V is the volume of the cube.

Formula to calculate the volume of the cube is,

$V=r^3$ (II)

From Equations (I) and (II),

$P=\frac{2}{3}\left(\frac{n{N}_A}{r^3}\right){\left(KE\right)}_{avg}$

Substitute 20 cm for r, $6.07\times {10}^{-21}\text{J}$ for ${\left(KE\right)}_{avg}$, 3 mol for n and $6.02\times {10}^{23}\text{atoms/mol}$ for $N_A$ to get P.

$\begin{array}{c}P=\frac{2}{3}\left[\frac{\left(3\text{mol}\right)\left(6.02\times {10}^{23}\text{atoms/mol}\right)}{{\left(20\text{cm}\right)}^3}\right]\left(6.07\times {10}^{-21}\text{J}\right)\\ =\frac{2}{3}\left[\frac{\left(3\text{mol}\right)\left(6.02\times {10}^{23}\text{atoms/mol}\right)}{{\left(20\times {10}^{-2}\text{m}\right)}^3}\right]\left(6.07\times {10}^{-21}\text{J}\right)\\ =9.14\times {10}^5\text{Pa}\end{array}$

Conclusion:

The gas pressure is $9.14\times {10}^5\text{Pa}$.

To determine

The gas pressure.

Answer to Problem 44P

The gas pressure is $9.13\times {10}^5\text{Pa}$.

Explanation of Solution

Given info: The number of moles (n) is 3. The temperature (T) is ${20}^{\text{ο}}\text{C}$. The side of the cube(r) is 20 cm. Molar mass of neon gas is $20.180\times {10}^{-3}\text{kg/mol}$.

From Ideal gas law, the gas pressure is,

$P=\frac{nRT}{V}$ (III)

Formula to calculate the volume of the cube is,

$V=r^3$ (IV)

From Equations (III) and (IV),

$P=\frac{nRT}{r^3}$

Substitute 20 cm for r, 3 mol for n ${20}^{\text{ο}}\text{C}$ for T and $8.31{\text{Jmol}}^{-1}{\text{K}}^{-1}$ for R to get P.

$\begin{array}{c}P=\frac{\left(3\text{mol}\right)\left(8.31{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\left({20}^{\text{ο}}\text{C}\right)}{{\left(20\text{cm}\right)}^3}\\ =\frac{\left(3\text{mol}\right)\left(8.31{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\left(20\text{+273}\text{K}\right)}{{\left(20\times {10}^{-2}\text{m}\right)}^3}\\ =9.13\times {10}^5\text{Pa}\end{array}$

Conclusion:

The gas pressure is $9.13\times {10}^5\text{Pa}$.