Question

The volume of a \(1-\mathrm{kg}\) copper sphere is not allowed to vary by more than \(0.1 \%\). If the pressure exerted on the sphere is increased from 10 bar while the temperature remains constant at \(300 \mathrm{~K}\), determine the maximum allowed pressure, in bar. Average values of \(\rho, \beta\), and \(\kappa\) are \(8888 \mathrm{~kg} / \mathrm{m}^{3}, 49.2 \times 10^{-6}\) \((\mathrm{K})^{-1}\), and \(0.776 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{N}\), respectively.

Short answer

1.289 x 10^7 bar

Step-by-step solution

- Understand the Given Values

Identify all the provided values:- Initial pressure, \(P_1 = 10 \text{ bar}\)- Temperature, \(T = 300 \text{ K}\)- Density of copper, \(\rho = 8888 \text{ kg/m}^3\)- Thermal expansion coefficient, \(\beta = 49.2 \times 10^{-6} \text{ K}^{-1}\)- Isothermal compressibility, \(\kappa = 0.776 \times 10^{-11} \text{ m}^2/\text{N}\)

- Set Up the Maximum Allowed Volume Change Condition

To find the maximum allowed pressure, we need to account for the maximum permissible volume change, which is \(0.1 \text{ percent}\).\(0.1 \text{ percent} = 0.001\)

- Relate Volume Change to Pressure Change

Using the relation between volume change and pressure:\(dV = V \cdot \kappa \cdot dP\)where \(dV/V = 0.001 \) and \(dP = P_2 - P_1\), solve for the maximum pressure \(P_2\).Given \( P_1 = 10 \text{ bar} \), \( \kappa = 0.776 \times 10^{-11} \text{ m}^{2} / \text{N} \), and \(0.001\), we have: \( 0.001 = \kappa (P_2 - 10 \text{ bar}) \)

- Calculate the Differential Pressure

Rearrange and solve for \(P_2\):\(0.001 = 0.776 \times 10^{-11} (P_2 - 10 \text{ bar})\)\(P_2 - 10 \text{ bar} = 0.001 / (0.776 \times 10^{-11})\)\(P_2 - 10 \text{ bar} = 1.289 \times 10^{7} \text{ bar}\)

- Determine Maximum Allowed Pressure

Finally, add the initial pressure to determine the maximum allowed pressure:\(P_2 = 1.289 \times 10^7 \text{ bar} + 10 \text{ bar}\)\( P_2 = 1.289 \times 10^7 \text{ bar}\)

Key concepts

Isothermal Compressibility

Isothermal compressibility, denoted as \(\kappa\), represents how much the volume of a material changes under pressure when the temperature remains constant. Mathematically, it is defined as:\[ \kappa = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T} \]Where \(V\) is the volume and \(P\) is the pressure. Essentially, it measures the material's resistance to compressive forces. The units are typically \( \text{m}^2/\text{N}\), which means square meters per newton. In our exercise, the compressibility of copper is given as \(0.776 \times 10^{-11} \text{m}^{2}/\text{N} \). This tiny value suggests copper is quite resistant to compression.

Thermal Expansion Coefficient

The thermal expansion coefficient \(\beta\) measures how much the volume of a material changes with temperature. It is defined as:\[ \beta = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P} \]Where \(V\) is the volume and \(T\) is the temperature. The units are usually \( \text{K}^{-1} \), indicating the rate of volume change per degree Kelvin. In this problem, the thermal expansion coefficient for copper is given as \(49.2 \times 10^{-6} \text{K}^{-1} \). This coefficient is crucial when dealing with problems involving temperature changes, but since the temperature remains constant here, thermal expansion does not directly affect our solution.

Volume Change Under Pressure

The volume change due to pressure is determined by isothermal compressibility. The relationship is given by:\[ dV = V \cdot \kappa \cdot dP \]Where \( dV \) is the change in volume, \( V \) is the initial volume, \( \kappa \) is the isothermal compressibility, and \( dP \) is the pressure change. In our exercise, the permissible volume change is \( 0.1\% = 0.001 \). Using this relationship and the known compressibility, we can derive the maximum allowable pressure change. Solving for the pressure,\[ \text{dP} = \frac{0.001}{0.776 \times 10^{-11}}= 1.289 \times 10^{7} \text{bar} \]Finally, adding the initial pressure, the maximum allowable pressure \( P_2 = 1.289 \times 10^7 \text{bar} + 10 \text{bar} = 1.289 \times 10^7 \text{bar} \).

Density of Materials

Density \( \rho \) measures the mass per unit volume of a substance. For copper in this problem, it is given as \(8888 \text{ kg/m}^3 \). Density plays a crucial role in understanding the mass and volume relationship of materials. Higher density means more mass in a given volume, which affects how substances react under different physical conditions like pressure and temperature. Though not directly used in our pressure calculation, knowing the density helps contextualize the behavior of materials under varying conditions and can be essential for related computations.