Chapter 4.5, Problem 54P

A mass of 10 g of nitrogen is contained in the spring-loaded piston–cylinder device shown in Fig. P4–54. The spring constant is 1 kN/m, and the piston diameter is 10 cm. When the spring exerts no force against the piston, the nitrogen is at 120 kPa and 27°C. The device is now heated until its volume is 10 percent greater than the original volume. Determine the change in the specific internal energy and enthalpy of the nitrogen.

FIGURE P4–54

To determine

The change in the internal energy of the nitrogen.

The change in the enthalpy of the nitrogen.

Answer to Problem 54P

The change in the internal energy of the nitrogen is $\underset{\_}{46.8\text{kJ}/\text{kg}}$.

The change in the enthalpy of the nitrogen is $\underset{\_}{65.5\text{kJ}/\text{kg}}$.

Explanation of Solution

Write the expression for the initial volume of nitrogen.

${\nu }_1=\frac{mR{T}_1}{P_1}$ (I)

Here, the mass of the spring loaded piston cylinder device is $m$, the universal gas constant is $R$, the initial temperature of the nitrogen gas is $T_1$, and the initial pressure of the nitrogen is $P_1$.

Determine the linear P-v process for spring loaded piston cylinder device.

$\begin{array}{c}P-P_1=\frac{F_s-F_{s,1}}{A}\\ =k\frac{x-x_1}{A}\\ =\frac{kA}{A^2}\left(x-x_1\right)\\ =\frac{k}{A^2}\left(\nu -{\nu }_1\right)\end{array}$

         $=c\left(\nu -{\nu }_1\right)$ (II)

Here, the system pressure is $P$, the initial pressure of system is $P_1$, the spring constant is $x$, the initial spring constant is $x_1$, the volume of the system is $\nu$, the initial volume of the system is ${\nu }_1$, the specific heat is $c$, and the area of the system is $A$.

Determine the specific heat constant value.

$\begin{array}{c}c=\frac{k}{A^2}\\ =\frac{4^{2}k}{{\pi }^2{D}^4}\end{array}$ (III)

Substitute $P_2$ for $P$ and ${\nu }_2$ for $\nu$ in Equation (II) to find out the final pressure of a spring loaded piston cylinder device.

$\begin{array}{l}{P}_2-P_1=c\left({\nu }_2-{\nu }_1\right)\\ P_2=P_1+c\left({\nu }_2-{\nu }_1\right)\end{array}$ (IV)

From the Equation (IV), the final volume is 10 percent greater than the original volume;$\left({\nu }_2=1.1{\nu }_1\right)$.

$\begin{array}{c}{P}_2=P_1+c\left(1.1{\nu }_1-{\nu }_1\right)\\ =P_1+0.1c{\nu }_1\end{array}$ (V)

Determine the final temperature of the nitrogen.

$\begin{array}{c}{T}_2=\frac{P_2{\nu }_2}{mR}\\ =\frac{P_2\left(1.1{\nu }_1\right)}{mR}\end{array}$ (VI)

Here, the final pressure of the nitrogen is $P_2$, the mass of the spring loaded piston cylinder device is $m$, the universal gas constant is $R$, and the final volume of the spring loaded piston cylinder device is ${\nu }_2$.

Determine the internal energy of the spring loaded piston-cylinder device.

$\begin{array}{c}\Delta u=c_{\nu }\Delta T\\ =c_{\nu }\left(T_2-T_1\right)\end{array}$ (VII)

Here, the specific heat of constant volume is $c_{\nu }$.

Determine the enthalpy of the spring loaded piston-cylinder device.

$\begin{array}{c}\Delta h=c_p\Delta T\\ =c_p\left(T_2-T_1\right)\end{array}$ (VIII)

Here, the specific heat of constant pressure is $c_p$.

Conclusion:

Write the conversion of unit for temperature of 20 C from $°\text{C}$ to K.

$\begin{array}{c}{T}_1=27°\text{C}\\ =\text{27}+273\text{K}\\ =\text{300}\text{K}\end{array}$

Refer Table A-2(a), “Ideal-gas specific heats of various common gases” to obtain the value of gas constant, specific heat of constant volume and pressure for nitrogen gas is $0.2968\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$, $0.743\text{kJ}/\text{kg}\cdot \text{K}$, and $1.039\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $10\text{g}$ for $m$, $0.2968\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$, $27°\text{C}$ for $T_1$, and 120 kPa for $P_1$ in Equation (I).

$\begin{array}{c}{\nu }_1=\frac{\left(10\text{g}\right)\left(0.2968\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(27°\text{C}\right)}{120\text{kPa}}\\ =\frac{\left(10\text{g}\times \left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(0.2968\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(27+273\text{K}\right)}{120\text{kPa}}\\ =\frac{\left(0.8904\text{kPa}\cdot {\text{m}}^3\right)}{120\text{kPa}}\\ =0.00742{\text{m}}^{\text{3}}\end{array}$

Substitute $1\text{kN}/\text{m}$ for $k$ and $10\text{cm}$ for $D$ in Equation (III).

$\begin{array}{c}c=\frac{4^2\left(1\text{kN}/\text{m}\right)}{{\pi }^2{\left(10\text{cm}\right)}^4}\\ =\frac{\left(16\right)\left(1\text{kN}/\text{m}\right)}{{\pi }^2{\left(10\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^4}\\ =\frac{16\text{kN}/\text{m}}{0.000987{\text{m}}^{\text{4}}}\\ =16,211\text{kN}/{\text{m}}^5\end{array}$

Substitute $120\text{kPa}$ for $P_1$, $16,211\text{kN}/{\text{m}}^{\text{5}}$ for $c$, and $0.00742{\text{m}}^{\text{3}}$ for ${\nu }_1$ in Equation (V).

$\begin{array}{c}{P}_2=120\text{kPa}+0.1\left(16,211\text{kN}/{\text{m}}^{\text{5}}\right)\left(0.00742{\text{m}}^{\text{3}}\right)\\ =120\text{kPa}+\left(1.20\text{kN}/{\text{m}}^{\text{2}}\times \left(\frac{1\text{kPa}}{1\text{kN}/{\text{m}}^{\text{2}}}\right)\right)\\ =132.0\text{kPa}\end{array}$

Substitute $132.0\text{kPa}$ for $P_2$, $0.00742{\text{m}}^{\text{3}}$ for ${\nu }_1$, 10 g for $m$, and $0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R$ in Equation (VI).

$\begin{array}{c}{T}_2=\frac{\left(132.0\text{kPa}\right)\left(1.1\times 0.00742{\text{m}}^{\text{3}}\right)}{\left(10\text{g}\right)\left(0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)}\\ =\frac{\left(1.077384\text{kPa}\cdot {\text{m}}^{\text{3}}\right)}{\left(10\text{g}\times \left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)}\\ =363\text{K}\end{array}$

Substitute $0.743\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{\nu }$, 363 K for $T_2$, 300 K for $T_1$ in Equation (VII).

$\begin{array}{c}\Delta u=\left(0.743\text{kJ}/\text{kg}\cdot \text{K}\right)\left(363-300\right)\text{K}\\ =\left(0.743\text{kJ}/\text{kg}\cdot \text{K}\right)\left(63\text{K}\right)\\ =46.8\text{kJ}/\text{kg}\end{array}$

Thus, the change in the internal energy of the nitrogen is $\underset{\_}{46.8\text{kJ}/\text{kg}}$.

Substitute $1.039\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 363 K for $T_2$, 300 K for $T_1$ in Equation (VII).

$\begin{array}{c}\Delta u=\left(1.039\text{kJ}/\text{kg}\cdot \text{K}\right)\left(363-300\right)\text{K}\\ =\left(1.039\text{kJ}/\text{kg}\cdot \text{K}\right)\left(63\text{K}\right)\\ =65.5\text{kJ}/\text{kg}\end{array}$

Thus, the change in the enthalpy of the nitrogen is $\underset{\_}{65.5\text{kJ}/\text{kg}}$.