Chapter 21, Problem 21.32P

Air (a diatomic ideal gas) at 27.0°C and atmospheric pressure is drawn into a bicycle pump that has a cylinder with an inner diameter of 2.50 cm and length 50.0 cm. The downstroke adiabatically compresses the air, which readies a gauge pressure of 8.00 × 10⁵ Pa before entering the tire. We wish to investigate the temperature increase of the pump. (a) What is the initial volume of the air in the pump? (b) What is the number of moles of air in the pump? (c) What is the absolute pressure of the compressed air? (d) What is the volume of the compressed air? (c) What is the temperature of the compressed air? (f) What is the increase in internal energy of the gas during the compression? What If? The pump is made of steel that is 2.00 mm thick. Assume 4.00 cm of the cylinder’s length is allowed to come to thermal equilibrium with the air. (g) What is the volume of steel in this 4.00-cm length? (h) What is the mass of steel in this 4.00-cm length? (i) Assume the pump is compressed once. After the adiabatic expansion, conduction results in the energy increase in part (f) being shared between the gas and the 4.00-cm length of steel. What will be the increase in temperature of the steel after one compression?

To determine

The initial volume of the air in the pump.

Answer to Problem 21.32P

The initial volume of the air in the pump is $2.45\times {10}^{-4}{\text{m}}^3$.

Explanation of Solution

Initial temperature for diatomic gasis $27.0°\text{C}$, length of the pump is $50.0\text{cm}$, diameter of the pump is $2.50\text{cm}$, gauge pressure for the compressed air is $8.00\times {10}^5\text{Pa}$.

Write the expression to calculate the radius of the pump.

$r=\frac{d}{2}$

Here,

$r$ is the radius of the pump.

$d$ is the diameter of the pump.

Write the formula to calculate the initial volume of the air in the pump.

$V_1=\pi r^{2}l$ (1)

Here,

$V_1$ is the initial volume of the air in the pump.

$l$ is the length of the pump.

Substitute $\frac{d}{2}$ for $r$ in equation (1) to find $V_1$,

    $V_1=\pi {\left(\frac{d}{2}\right)}^{2}l$ (2)

Substitute $2.50\text{cm}$ for $d$, $50.0\text{cm}$ for $l$ in equation (2) to find $V_1$,

    $\begin{array}{c}{V}_1=\pi {\left(\frac{2.50\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)}{2}\right)}^2\left(50.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =2.45\times {10}^{-4}{\text{m}}^3\end{array}$

Thus, the initial volume of the air in the pump is $2.45\times {10}^{-4}{\text{m}}^3$.

Conclusion:

Therefore, the initial volume of the air in the pump is $2.45\times {10}^{-4}{\text{m}}^3$.

To determine

The number of moles of air in the pump.

Answer to Problem 21.32P

The number of moles of air in the pump is $9.97\times {10}^{-3}\text{moles}$.

Explanation of Solution

Initial temperature for diatomic gas is $27.0°\text{C}$, length of the pump is $50.0\text{cm}$, diameter of the pump is $2.50\text{cm}$, gauge pressure for the compressed air is $8.00\times {10}^5\text{Pa}$.

Write the formula to calculate the number of moles of air in the pump.

    $\begin{array}{c}{P}_1{V}_1=nR{T}_1\\ n=\frac{P_1{V}_1}{R{T}_1}\end{array}$        (3)

Here,

$n$ is the number of moles of air in the pump.

$P_1$ is the atmospheric pressure for diatomic gas.

$R$ is the ideal gas constant.

$T_1$ is the initial temperature for diatomic gas.

The value of atmospheric pressure for diatomic gas is $1.013\times {10}^5\text{Pa}$ and the value of ideal gas constant is $8.314\text{J}/\text{molK}$.

Substitute $2.45\times {10}^{-4}{\text{m}}^3$ for $V_1$, $1.013\times {10}^5\text{Pa}$ for $P_1$, $8.314\text{J}/\text{molK}$ for $R$, $27.0°\text{C}$ for $T_1$ in equation (3) to find $n$,

    $\begin{array}{c}n=\frac{\left(1.013\times {10}^5\text{Pa}\right)\times \left(2.45\times {10}^{-4}{\text{m}}^3\right)}{\left(8.314\text{J}/\text{molK}\right)\times \left(27.0°\text{C}+273\right)\text{K}}\\ =9.97\times {10}^{-3}\text{moles}\end{array}$

Thus, the number of moles of air in the pump is $9.97\times {10}^{-3}\text{moles}$.

Conclusion:

Therefore, the number of moles of air in the pump is $9.97\times {10}^{-3}\text{moles}$.

To determine

The absolute pressure of the compressed air.

Answer to Problem 21.32P

The absolute pressure of the compressed air is $9.01\times {10}^5\text{Pa}$.

Explanation of Solution

Initial temperature for diatomic gas is $27.0°\text{C}$, length of the pump is $50.0\text{cm}$, diameter of the pump is $2.50\text{cm}$, gauge pressure for the compressed air is $8.00\times {10}^5\text{Pa}$.

Write the formula to calculate the absolute pressure of the compressed air.

    $P_2=P_g+P_1$ (4)

Here,

$P_2$ is the absolute pressure of the compressed air.

$P_g$ is the gauge pressure for the compressed air.

Substitute $8.00\times {10}^5\text{Pa}$ for $P_g$, $1.013\times {10}^5\text{Pa}$ for $P_1$ in equation (4) to find $P_2$,

    $\begin{array}{c}{P}_2=8.00\times {10}^5\text{Pa}+1.013\times {10}^5\text{Pa}\\ =9.013\times {10}^5\text{Pa}\\ \approx 9.01\times {10}^5\text{Pa}\end{array}$

Thus, the absolute pressure of the compressed air is $9.01\times {10}^5\text{Pa}$.

Conclusion:

Therefore, the absolute pressure of the compressed air is $9.01\times {10}^5\text{Pa}$.

To determine

The volume of the compressed air.

Answer to Problem 21.32P

The volume of the compressed air is $5.15\times {10}^{-5}{\text{m}}^3$.

Explanation of Solution

Initial temperature for diatomic gas is $27.0°\text{C}$, length of the pump is $50.0\text{cm}$, diameter of the pump is $2.50\text{cm}$, gauge pressure for the compressed air is $8.00\times {10}^5\text{Pa}$.

Write the expression for the adiabatic compression.

    $P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

Here,

$V_2$ is the volume of the compressed air.

$\gamma$ is the specific heat ratio.

Write the formula to calculate the volume of the compressed air.

    $V_2=V_1{\left(\frac{P_1}{P_2}\right)}^{\frac{1}{\gamma }}$ (5)

Substitute $9.01\times {10}^5\text{Pa}$ for $P_2$, $1.013\times {10}^5\text{Pa}$ for $P_1$, $2.45\times {10}^{-4}{\text{m}}^3$ for $V_1$, $1.40$ for $\gamma$ in equation (5) to find $V_2$,

    $\begin{array}{c}{V}_2=\left(2.45\times {10}^{-4}{\text{m}}^3\right){\left(\frac{1.013\times {10}^5\text{Pa}}{9.01\times {10}^5\text{Pa}}\right)}^{\frac{1}{1.40}}\\ =5.145\times {10}^{-5}{\text{m}}^3\\ \approx 5.15\times {10}^{-5}{\text{m}}^3\end{array}$

Thus, the volume of the compressed air is $5.15\times {10}^{-5}{\text{m}}^3$.

Conclusion:

Therefore, the volume of the compressed air is $5.15\times {10}^{-5}{\text{m}}^3$.

To determine

The temperature of the compressed air.

Answer to Problem 21.32P

The temperature of the compressed air is $560\text{K}$.

Explanation of Solution

Initial temperature for diatomic gas is $27.0°\text{C}$, length of the pump is $50.0\text{cm}$, diameter of the pump is $2.50\text{cm}$, gauge pressure for the compressed air is $8.00\times {10}^5\text{Pa}$.

Write the formula to calculate the temperature of the compressed air.

    $\begin{array}{c}{P}_2{V}_2=nR{T}_2\\ T_2=\frac{P_2{V}_2}{nR}\end{array}$        (6)

Here,

$T_2$ is the temperature of the compressed air.

Substitute $5.15\times {10}^{-5}{\text{m}}^3$ for $V_2$, $9.013\times {10}^5\text{Pa}$ for $P_2$, $8.314\text{J}/\text{molK}$ for $R$, $9.97\times {10}^{-3}\text{moles}$ for $n$ in equation (6) to find $T_2$,

    $\begin{array}{c}{T}_2=\frac{P_2{V}_2}{nR}\\ T_2=\frac{\left(9.013\times {10}^5\text{Pa}\right)\times \left(5.15\times {10}^{-5}{\text{m}}^3\right)}{\left(9.97\times {10}^{-3}\text{moles}\right)\times \left(8.314\text{J}/\text{molK}\right)}\\ =559.978\text{K}\\ \approx 560\text{K}\end{array}$

Thus, the temperature of the compressed air is $560\text{K}$.

Conclusion:

Therefore, the temperature of the compressed air is $560\text{K}$.

To determine

The increase in internal energy of the gas during the compression.

Answer to Problem 21.32P

The increase in internal energy of the gas during the compression is $53.9\text{J}$.

Explanation of Solution

Initial temperature for diatomic gas is $27.0°\text{C}$, length of the pump is $50.0\text{cm}$, diameter of the pump is $2.50\text{cm}$, gauge pressure for the compressed air is $8.00\times {10}^5\text{Pa}$.

For adiabatic process, the work done on the gas is equal to the change in internal energyof the gas during the compression.

    $W=\Delta U$ (7)

Here,

$W$ is the work done on the gas.

$\Delta U$ is the change in internal energyof the gas during the compression.

Write the expression for the change in internal energyof the gas during the compression.

    $\Delta U=n{C}_V\Delta T$ (8)

Here,

$C_V$ is the specific heat at constant volume.

Write the expression for specific heat at constant volume.

    $C_V=\frac{5}{2}R$ (9)

Here,

$C_V$ is the specific heat at constant volume.

Equate the three expressions (7),(8) and (9)and re-arrange to get $\Delta U$,

    $\Delta U=n\left(\frac{5}{2}R\right)\Delta T$ (10)

Write the formula to calculate the change in temperature of a monatomic ideal gas.

    $\Delta T=T_2-T_1$ (11)

Here,

$\Delta T$ is the change in temperature of a monatomic ideal gas.

Substitute $560\text{K}$ for $T_2$, $300\text{K}$ for $T_1$ in equation (11) to find $\Delta T$,

    $\begin{array}{c}\Delta T=T_2-T_1\\ =560\text{K}-300\text{K}\\ =260\text{K}\end{array}$

Thus, the change in temperature of a monatomic ideal gas is $260\text{K}$.

Substitute $260\text{K}$ for $\Delta T$, $8.314\text{J}/\text{molK}$ for $R$, $9.97\times {10}^{-3}\text{moles}$ for $n$ in equation (10) to find $\Delta U$,

    $\begin{array}{c}\Delta U=\left(9.97\times {10}^{-3}\text{moles}\right)\left(\frac{5}{2}\times 8.314\text{J}/\text{molK}\right)\times 260\text{K}\\ \text{=53}\text{.87}\text{J}\\ \approx \text{53}\text{.9}\text{J}\end{array}$

Thus, the increase in internal energy of the gas during the compression is $53.9\text{J}$.

Conclusion:

Therefore, the increase in internal energy of the gas during the compression is $53.9\text{J}$.

To determine

The volume of the steel in this $4.00\text{cm}$ length.

Answer to Problem 21.32P

The volume of the steel in this $4.00\text{cm}$ length is $6.79\times {10}^{-6}{\text{m}}^3$.

Explanation of Solution

Initial temperature for diatomic gas is $27.0°\text{C}$, length of the pump is $50.0\text{cm}$, length of the cylinder is $4.00\text{cm}$, thickness of the pump is $2.00\text{mm}$, diameter of the pump is $2.50\text{cm}$, gauge pressure for the compressed air is $8.00\times {10}^5\text{Pa}$.

Write the formula to calculate the volume of the steel in this $4.00\text{cm}$ length.

    $V_s=\pi r_s^{2}l$ (12)

Here,

$r_s$ is the radius of the pump for steel.

$V_s$ is the volume of the steel in this $4.00\text{cm}$ length.

Write the expression to calculate the square radius of the pump for steel.

    $r_s^2=r_2^2-{\left(\frac{d}{2}\right)}^2$ (13)

Here,

$r_2$ is the outer radius of the pump.

Write the formula to calculate the outer radius of the pump.

    $r_2=\frac{d}{2}+t$ (14)

Here,

$t$ is the thickness of the pump.

Substitute $2.00\text{mm}$ for $t$, $2.50\text{cm}$ for $r$ in equation (14) to find $r_2$,

    $\begin{array}{c}{r}_2=\left(\frac{2.50\text{cm}}{2}\times \frac{10\text{mm}}{1\text{cm}}\right)+\left(2.00\text{mm}\right)\\ =14.5\text{mm}\end{array}$

Thus, the outer radius of the pump is $14.5\text{mm}$.

Substitute $14.5\text{mm}$ for $r_2$, $2.50\text{cm}$ for $r$ in equation (13) to find $r_s$,

    $\begin{array}{c}{r}_s^2=r_2^2-{\left(\frac{d}{2}\right)}^2\\ ={\left(14.5\text{mm}\right)}^2-{\left(\frac{2.5\text{cm}\times \frac{10\text{mm}}{1\text{cm}}}{2}\right)}^2\\ =54\text{mm}\end{array}$

Thus, the square radius of the pump for steel is $54\text{mm}$.

Substitute $54\text{mm}$ for $r_s^2$, $4.00\text{cm}$ for $l$ in equation (12) to find $V_s$,

    $\begin{array}{c}{V}_s=\pi \times \left(54\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\times \left(4.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =6.785\times {10}^{-6}{\text{m}}^3\\ \approx 6.79\times {10}^{-6}{\text{m}}^3\end{array}$

Thus, the volume of the steel in this $4.00\text{cm}$ length is $6.79\times {10}^{-6}{\text{m}}^3$.

Conclusion:

Therefore, the volume of the steel in this $4.00\text{cm}$ length is $6.79\times {10}^{-6}{\text{m}}^3$.

To determine

The mass of the steel in this $4.00\text{cm}$ length.

Answer to Problem 21.32P

The mass of the steel in this $4.00\text{cm}$ length is $53.3\text{g}$.

Explanation of Solution

Initial temperature for diatomic gas is $27.0°\text{C}$, length of the pump is $50.0\text{cm}$, length of the cylinder is $4.00\text{cm}$, thickness of the pump is $2.00\text{mm}$, diameter of the pump is $2.50\text{cm}$, gauge pressure for the compressed air is $8.00\times {10}^5\text{Pa}$.

Write the formula to calculate the mass of the steel in this $4.00\text{cm}$ length.

    $\begin{array}{c}{\rho }_s=\frac{m}{V_s}\\ m={\rho }_s\times V_s\end{array}$ (15)

Here,

$m$ is the mass of the steel in this $4.00\text{cm}$ length.

${\rho }_s$ is the density of the steel.

The value of density of the steel is $7.85\times {10}^6\text{g}/{\text{m}}^{\text{3}}$.

Substitute $7.85\times {10}^6\text{g}/{\text{m}}^{\text{3}}$ for ${\rho }_s$, $6.79\times {10}^{-6}{\text{m}}^3$ for $V_s$, in equation (15) to find $m$,

    $\begin{array}{c}m=\left(7.85\times {10}^6\text{g}/{\text{m}}^{\text{3}}\right)\times \left(6.79\times {10}^{-6}{\text{m}}^3\right)\\ =53.3\text{g}\end{array}$

Thus, the mass of the steel in this $4.00\text{cm}$ length is $53.3\text{g}$.

Conclusion:

Therefore, the mass of the steel in this $4.00\text{cm}$ length is $53.3\text{g}$.

To determine

The increase in temperature of the steel after one compression.

Answer to Problem 21.32P

The increase in temperature of the steel after one compression is $2.24\text{K}$.

Explanation of Solution

Initial temperature for diatomic gas is $27.0°\text{C}$, length of the pump is $50.0\text{cm}$, length of the cylinder is $4.00\text{cm}$, thickness of the pump is $2.00\text{mm}$, diameter of the pump is $2.50\text{cm}$, gauge pressure for the compressed air is $8.00\times {10}^5\text{Pa}$.

After the adiabatic compression, conduction in the part (f) being shared between the gas and the $4.00\text{cm}$ length of steel.

The work done on the gas is equal to the sum of change in internal energyof the gas during the compression and the heat supplied.

    $W=\Delta U+Q$ (16)

Here,

$Q$ is the heat supplied during the compression which is zero for adiabatic compression.

Write the expression for $\Delta U$ of both the cylinder and the steel.

  $\Delta U=n{C}_V\Delta T+mC\Delta T$ (17)

Here,

$C$ is the specific heat capacity of steel in $\text{J}/\text{kg}\cdot \text{K}$.

The value of specific heat capacity is $447\text{J}/\text{kg}\cdot \text{K}$.

Substitute $0$ for $Q$ and the three expressions (8),(9),(16) in (17), then re-arrange to get $\Delta T$,

    $\begin{array}{c}W=n{C}_V\Delta T+mC\Delta T+0\\ =\Delta T\left(n\left(\frac{5}{2}R\right)+mC\right)\\ \Delta T=\frac{W}{\left(n\left(\frac{5}{2}R\right)+mC\right)}\end{array}$ (18)

Substitute $53.9\text{J}$ for $W$, $8.314\text{J}/\text{molK}$ for $R$, $9.97\times {10}^{-3}\text{moles}$ for $n$, $53.3\text{g}$ for $m$, $447\text{J}/\text{kg-}\text{K}$ for $C$ in equation (17) to find $\Delta U$,

    $\begin{array}{c}\Delta T=\frac{\text{53}\text{.9}\text{J}}{\left(\left(9.97\times {10}^{-3}\text{moles}\right)\left(\frac{5}{2}\times 8.314\text{J}/\text{molK}\right)+\left(53.3\text{g}\times \frac{1\text{kg}}{1000\text{g}}\right)\times \left(447\text{J}/\text{kg}\cdot \text{K}\right)\right)}\\ \Delta T=\frac{\text{53}\text{.9}\text{J}}{\left(0.20722+23.55\right)}\\ =2.242\text{K}\\ \approx 2.24\text{K}\end{array}$

Thus, the increase in temperature of the steel after one compression is $2.24\text{K}$.

Conclusion:

Therefore, the increase in temperature of the steel after one compression is $2.24\text{K}$.