Chapter 21, Problem 32P

(a)

To determine

The initial volume of the air in the pump.

Answer to Problem 32P

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

Explanation of Solution

Write the expression for volume of the cylinder.

  $V_i=\pi r^{2}h$

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

Rewrite the above equation.

  $V_i=\pi {\left(\frac{d}{2}\right)}^{2}h$                                                                                                     (I)

Here, $d$ is the diameter of the bicycle cylinder pump.

Conclusion:

Substitute $2.50\text{cm}$ for $d$ and $50\text{cm}$ for $h$ in equation (I) to find $V_i$.

  $\begin{array}{c}{V}_i=3.14{\left(\frac{2.50\text{cm}}{2}\right)}^2\left(50\text{cm}\right)\\ =3.14{\left[\left(1.25\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\right]}^2\left(50\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\\ =2.45\times {10}^{-4}{\text{m}}^3\end{array}$

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

(b)

To determine

The number of moles of air in the pump.

Answer to Problem 32P

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

Explanation of Solution

Write the expression from ideal gas law.

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

Here, $n$ is the number of moles, $P_i$ is the initial atmospheric pressure, $V_i$ is the initial volume, $R$ is the gas constant, and $T_i$ is the initial temperature.

Conclusion:

Substitute $1.013\times {10}^5\text{Pa}$ for $P_i$, $2.45\times {10}^{-4}{\text{m}}^3$ for $V_i$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, and $27.0°\text{C}$ for $T_i$ in equation (II) to find $n$.

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

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

(c)

To determine

The absolute pressure of the compressed air.

Answer to Problem 32P

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

Explanation of Solution

Write the expression for absolute pressure.

  $P_{\text{abs}}=P_{\text{g}}+P_{\text{e}}$                                                                                                    (III)

Here, $P_{\text{abs}}$ is the absolute pressure, $P_{\text{g}}$ is the gauge pressure, and $P_{\text{e}}$ is the external pressure.

Conclusion:

Substitute $101.3\text{kPa}$ for $P_{\text{g}}$ and $800\text{kPa}$ for $P_{\text{e}}$ in equation (III) to find $P_{\text{abs}}$.

  $\begin{array}{c}{P}_{\text{abs}}=101.3\text{kPa}+800\text{kPa}\\ =901.3\text{kPa}\left(\frac{{10}^3\text{Pa}}{1\text{kPa}}\right)\\ =901.3\times {10}^3\text{Pa}\\ \approx 9.01\times {10}^5\text{Pa}\end{array}$

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

(d)

To determine

The volume of the compressed air.

Answer to Problem 32P

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

Explanation of Solution

Write the expression from an adiabatic compression process.

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

Here, $P_i$ is the initial pressure, $P_f$ is the final pressure, $V_i$ is the initial volume, $\gamma$ is the adiabatic compression, and $V_f$ is the final volume.

Rearrange the above expression for $V_f$.

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

Conclusion:

Substitute $2.45\times {10}^{-4}{\text{m}}^3$ for $V_i$, $901.3\times {10}^3\text{Pa}$ for $P_f$, $101.3\times {10}^3\text{Pa}$ for $P_i$, and $5/7$ for $1/\gamma$ in equation (IV) to find $V_f$.

  $\begin{array}{c}{V}_f=\left(2.45\times {10}^{-4}{\text{m}}^3\right){\left(\frac{101.3\times {10}^3\text{Pa}}{901.3\times {10}^3\text{Pa}}\right)}^{5/7}\\ =5.15\times {10}^{-5}{\text{m}}^3\end{array}$

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

(e)

To determine

The temperature of the compressed air.

Answer to Problem 32P

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

Explanation of Solution

Write the expression from an adiabatic compression process to find final temperature.

  $T_f=T_i\frac{P_f{V}_f}{P_i{V}_i}$

Here, $T_i$ is the initial temperature, $P_f$ is the final pressure, and $V_f$ is the final volume.

Rewrite the above expression.

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

Conclusion:

Substitute $27.0°\text{C}$ for $T_i$, $901.3\times {10}^3\text{Pa}$ for $P_f$, $101.3\times {10}^3\text{Pa}$ for $P_i$, and $5/7$ for $1/\gamma$ in equation (V) to find $T_f$.

  $\begin{array}{c}{T}_f=\left(27.0+273\text{K}\right){\left(\frac{101.3\times {10}^3\text{Pa}}{901.3\times {10}^3\text{Pa}}\right)}^{\left(5/7-1\right)}\\ =560\text{K}\end{array}$

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

(f)

To determine

The increasing internal energy of the gas.

Answer to Problem 32P

The increasing internal energy of the gas is $53.9\text{J}$.

Explanation of Solution

Write the expression for internal energy of the any gas.

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

Here, $\Delta E_{\mathrm{int}}$ is the internal energy of the any gas and $C_V$ is the specific heat capacity at constant volume.

The specific heat capacity at constant volume for diatomic gas is,

  $C_V=\frac{5}{2}R$                                                                                                        (VII)

Conclusion:

Substitute equation (VII) in equation (VI).

  $\Delta E_{\mathrm{int}}=\frac{5}{2}nR\Delta T$

Substitute $9.97\times {10}^{-3}\text{mol}$ for $n$, $560\text{K}-300\text{K}$ for $\Delta T$, and $8.314\text{J/mol}\cdot \text{K}$ for $R$ in above equation to find $\Delta E_{\mathrm{int}}$.

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

Therefore, the increasing internal energy of the gas is $53.9\text{J}$.

(g)

To determine

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

Answer to Problem 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

Write the expression for volume of the steel.

  $V=\left[\pi r_{\text{in}}^2-\pi r_{\text{out}}^2\right]h$                                                                                             (VIII)

Here, $r_{\text{in}}$ is the inner radius, $r_{\text{out}}$ is the outer radius, and $h$ is the length of the steel bar.

Conclusion:

Substitute $14.5\times {10}^{-3}\text{m}$ for $r_{\text{in}}$, $12.5\times {10}^{-3}\text{m}$ for $r_{\text{out}}$, and $4.00\text{cm}$ for $h$ in equation (VIII) to find $V$.

  $\begin{array}{c}V=3.14\left[{\left(14.5\times {10}^{-3}\text{m}\right)}^2-{\left(12.5\times {10}^{-3}\text{m}\right)}^2\right]\left(4.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =6.79\times {10}^{-6}{\text{m}}^3\end{array}$

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

(h)

To determine

The mass of the steel.

Answer to Problem 32P

The mass of the steel is $53.3\text{g}$.

Explanation of Solution

Write the expression for mass of the steel.

  $m=\rho V$                                                                                              (IX)

Here, $m$ is the mass of the steel, $\rho$ is the density of the steel, and $V$ is the volume 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 equation (IX) to find $m$.

  $\begin{array}{c}m=\left(7.86\times {10}^3{\text{kg/m}}^3\right)\left(6.79\times {10}^{-6}{\text{m}}^3\right)\\ =53.3\times {10}^{-3}\text{kg}\left(\frac{{10}^3\text{g}}{1\text{kg}}\right)\\ =53.3\text{g}\end{array}$

Therefore, the mass of the steel is $53.3\text{g}$.

(i)

To determine

The increasing temperature of the steel.

Answer to Problem 32P

The increasing temperature of the steel is $2.24\text{K}$.

Explanation of Solution

Write the expression for temperature of the steel.

  $\begin{array}{c}W=n{C}_V\Delta T+mc\Delta T\\ \Delta T=\frac{W}{n{C}_V+mc}\end{array}$                                                                                               (X)

Here, $W$ is the work done and $c$ is the specific heat capacity of the steel.

Rewrite the above expression.

  $\Delta T=\frac{W}{\frac{5}{2}nR+mc}$

Conclusion:

Substitute $53.9\text{J}$ for $W$, $9.97\times {10}^{-3}\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $448\text{J/kg}\cdot °\text{C}$ for $c$, and $53.3\times {10}^{-3}\text{kg}$ for $m$ in above equation to find $\Delta T$.

  $\begin{array}{c}\Delta T=\frac{53.9\text{J}}{\frac{5}{2}\left(9.97\times {10}^{-3}\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)+\left(53.3\times {10}^{-3}\text{kg}\right)\left(448\text{J/kg}°\text{C}\right)}\\ =2.24°\text{C}\\ \approx \text{2}\text{.24}\text{K}\end{array}$

Therefore, the increasing temperature of the steel is $2.24\text{K}$.