Chapter 17, Problem 53P

Air (a diatomic ideal gas) at 27.0°C and atmospheric pressure is drawn into a bicycle pump (Figure P17.53) that has a cylinder with an inner diameter of 2.50 cm and length 50.0 cm. The downstroke adiabatically compresses the air, which reaches 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? (e) 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?

Figure P17.53

To determine

The initial volume of the air in the pump.

Answer to Problem 53P

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

Explanation of Solution

Write the expression for the initial volume of the air in the pump,

    $V_i=\pi r^{2}l$        (I)

Here, $V_i$ is the initial volume of the air in the pump, $r$ is the radius of the pump and $l$ is the length of the pump.

Conclusion:

Substitute $\frac{2.50\times {10}^{-2}}{2}\text{m}$ for $r$ and $0.50\text{m}$ for $l$ in (I) to find $V_i$,

    $\begin{array}{c}{V}_i=\pi \left(\frac{2.50\times {10}^{-2}}{2}\text{m}\right)\left(0.50\text{m}\right)\\ =2.45\times {10}^{-4}{\text{m}}^3\end{array}$

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

To determine

The number of miles of the air in the pump .

Answer to Problem 53P

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

Explanation of Solution

Write the expression for the ideal gas law,

    $P_i{V}_i=nR{T}_i$        (II)

Here, $P_i$ is the initial pressure of the air, $R$ is the universal gas constant, $n$ is the number of moles of air and $C_V$ is the molar specific heat at constant volume.

Rewrite the above expression for $n$ ,

    $n=\frac{P_i{V}_i}{R{T}_i}$        (III)

Conclusion:

Substitute $1.013\times {10}^{5}Pa$ for $P_i$, $2.45\times {10}^{-4}{\text{m}}^3$ for $V_i$, $300\text{K}$ for $T_i$ and $8.314\text{J/mol}\text{.K}$ for $R$ in (III) to find $n$,

    $\begin{array}{c}n=\frac{\left(1.013\times {10}^{5}Pa\right)\left(2.45\times {10}^{-4}{\text{m}}^3\right)}{\left(8.314\text{J/mol}\text{.K}\right)\left(300\text{K}\right)}\\ =9.97\times {10}^{-3}\text{mol}\end{array}$

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

To determine

The absolute pressure of the compressed air .

Answer to Problem 53P

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

Explanation of Solution

Write the expression for the absolute pressure of the compressed air,

    $P_f=P_g+P_{ext}$        (IV)

Here, $P_f$ is the absolute pressure of the compressed air, $P_g$ is the gauge pressure and $P_{ext}$ is the external pressure.

Conclusion:

Substitute $101.3\text{kPa}$ for $P_{ext}$ and $800\text{kPa}$ for $P_g$ in (IV) to find $P_f$,

    $\begin{array}{c}{P}_f=101.3\text{kPa}+800\text{kPa}\\ =901.3\text{kPa}\\ \text{=9}\text{.01}\times {\text{10}}^5\text{kPa}\end{array}$

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

To determine

The volume of the compressed air .

Answer to Problem 53P

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

Explanation of Solution

Write the expression for the adiabatic compression,

    $P_i{V}_i{}^{\gamma }=P_f{V}_f{}^{\gamma }$        (V)

Here, $P_f$ is the final pressure of the compressed air, $P_i$ is the initial pressure of the air, $V_i$ is the initial volume of air, $V_f$ is the final volume of the compressed air and $\gamma$ is the ratio of the molar specific heat.

Rewrite the above expression for $V_f$,

    $V_f=V_i{\left(\frac{P_i}{P_f}\right)}^{1/\gamma }$        (VI)

Conclusion:

Substitute $101.3\text{kPa}$ for $P_i$, $901.3\text{kPa}$ for $P_f$, $7/5$ for $\gamma$ and $2.45\times {10}^{-4}{\text{m}}^{\text{3}}$ for $V_i$ in (VI) to find $V_f$ ,

    $\begin{array}{c}{V}_f=\left(2.45\times {10}^{-4}{\text{m}}^{\text{3}}\right)\left(\frac{101.3\text{kPa}}{901.3\text{kPa}}\right)\\ =5.15\times {10}^{-5}{\text{m}}^{\text{3}}\end{array}$

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

To determine

The temperature of the compressed air .

Answer to Problem 53P

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

Explanation of Solution

Write the expression for the ideal gas law for the compressed air,

    $P_f{V}_f=nR{T}_f$        (VII)

Here, $P_f$ is the absolute pressure of the compressed air, $V_f$ is the volume of compressed air, $n$ is the number of molecule, $R$ is the universal gas constant and $T_f$ is the temperature of the compressed air.

Rewrite the above equation for $T_f$,

    $\begin{array}{c}{T}_f=T_i\left(\frac{P_f{V}_f}{P_i{V}_i}\right)\\ =T_i\frac{P_f}{P_i}{\left(\frac{P_i}{P_f}\right)}^{1/\gamma }\\ =T_i{\left(\frac{P_i}{P_f}\right)}^{\left(1/\gamma -1\right)}\end{array}$        (VIII)

Conclusion:

Substitute $101.3\text{kPa}$ for $P_i$, $901.3\text{kPa}$ for $P_f$, $7/5$ for $\gamma$ and $300\text{K}$ for $T_i$ in (VIII) to find $T_f$,

    $\begin{array}{c}{T}_f=\left(300\text{K}\right){\left(\frac{101.3\text{kPa}}{901.3\text{kPa}}\right)}^{\left(5/7-1\right)}\\ =560\text{K}\end{array}$

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

To determine

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

Answer to Problem 53P

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

Explanation of Solution

Write the expression for the work done on the gas,

    $W=\Delta E_{\mathrm{int}}$        (IX)

Here, $W$ is the work done on the gas and $\Delta E_{\mathrm{int}}$ is the increase in internal energy of the gas.

Write the expression for the increase in internal energy of the gas,

    $\Delta E_{\mathrm{int}}=n{C}_V\Delta T$        (X)

Here, $C_V$ is the molar specific heat at constant volume and $\Delta T$ is change in temperature.

Conclusion:

Substitute $9.97\times {10}^{-3}\text{mol}$ for $n$, $\frac{5}{2}R$, $8.314\text{J/mol}\text{.K}$ for $R$ and $560\text{K}-300\text{K}$ for $\Delta T$ in (X) to find $\Delta E_{\mathrm{int}}$,

    $\begin{array}{c}\Delta E_{\mathrm{int}}=\left(9.97\times {10}^{-3}\text{mol}\right)\frac{5}{2}\left(8.314\text{J/mol}\text{.K}\right)\left(560\text{K}-300\text{K}\right)\\ \text{=}53.9\text{J}\end{array}$

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

To determine

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

Answer to Problem 53P

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

Explanation of Solution

Write the expression for the volume of steel,

    $V_s=\left(V_o-V_i\right)l$        (XI)

Here, $V_s$ is the volume of the steel, $V_o$ is the outer volume of the steel, $V_i$ is the inner side volume of the steel and $l$ is the length of the steel.

Write the expression for the volume of the steel,

    $V=\pi r^{2}l$        (XII)

Here, $r$ is the radius of the steel

Rewrite equation (XI),

    $V_s=\left(\pi r_o{}^2-\pi r_i{}^2\right)l$        (XIII)

Here, $r_o$  is the radius of the outer side steel and $r_i$ is the radius of the inner side steel.

Conclusion:

Substitute $14.5\times {10}^{-3}\text{m}$ for $r_o$, $12.5\times {10}^{-3}\text{m}$ for $r_i$, $4.00\times {10}^{-2}\text{m}$ for $l$ in (XIII) to find $V$,

    $\begin{array}{c}V=\left[\pi \left(14.5\times {10}^{-3}\text{m}\right)-\pi \left(12.5\times {10}^{-3}\text{m}\right)\right]\times \left(4.00\times {10}^{-2}\text{m}\right)\\ =6.79\times {10}^{-6}{\text{m}}^3\end{array}$

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

To determine

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

Answer to Problem 53P

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

Explanation of Solution

Write the expression for the mass of the steel,

    $m=\rho V$        (XIV)

Here, $m$ is the mass of the steel and $\rho$ is the density of the steel.

Conclusion:

Substitute $7.86\times {10}^3{\text{kg/m}}^3$ for $\rho$ and $6.79\times {10}^{-6}{\text{m}}^3$ for $V$ in (XIV) to find $m$ ,

    $\begin{array}{c}m=\left(7.86\times {10}^3{\text{kg/m}}^3\right)\left(6.79\times {10}^{-6}{\text{m}}^3\right)\\ =53.3\text{g}\end{array}$

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

To determine

The increase in temperature of the steel after onecompression .

Answer to Problem 53P

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

Explanation of Solution

From part (f),

Write the expression for the increase in internal energy of the gas,

    $\Delta E_{\mathrm{int}}=n{C}_V\Delta T+mc\Delta T$        (XV)

Here, $m$ is the mass of the steel, $c$ is the specific heat capacity, $C_V$ is the molar specific heat at constant volume and $\Delta T$ is change in temperature.

Rewrite the above expression,

    $\Delta T=\frac{\Delta E_{\mathrm{int}}}{n{C}_V+mc}$        (XVI)

Conclusion:

Substitute $53.9\text{J}$ for $\Delta E_{\mathrm{int}}$, $9.97\times {10}^{-3}\text{mol}$ for $n$, $8.314\text{J/mol}\text{.K}$ for $R$, $\frac{5}{2}R$ for $C_V$, $448\text{J/kg}{\text{.}}^{\text{o}}\text{C}$ and $0.0533\text{kg}$ for $m$ in (XVI) to find $\Delta T$ ,

    $\begin{array}{c}\Delta T=\frac{53.9\text{J}}{\left(9.97\times {10}^{-3}\text{mol}\right)\frac{5}{2}\left(8.314\text{J/mol}\text{.K}\right)+\left(0.0533\text{kg}\right)\left(448\text{J/kg}{\text{.}}^{\text{o}}\text{C}\right)}\\ =2.24{}^{\text{o}}\text{C}\\ \text{=}2.24\text{K}\end{array}$

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