Thermal Expansion for fluid Questions in English

(B) The volume of mercury that overflows is equal to the difference in the volume expansion of mercury and the glass flask.

$\Delta V = V_0 (\gamma_m - \gamma_g) \Delta \theta$

Given that $\gamma_g = 3 \alpha_g$, where $\alpha_g$ is the coefficient of linear expansion of glass.

$\gamma_g = 3 \times (0.1 \times 10^{-4} /{ }^{\circ} C) = 0.3 \times 10^{-4} /{ }^{\circ} C = 30 \times 10^{-6} /{ }^{\circ} C$

Given $\gamma_m = 1.82 \times 10^{-4} /{ }^{\circ} C = 182 \times 10^{-6} /{ }^{\circ} C$, $V_0 = 1 \text{ litre} = 1000 \text{ cc}$, and $\Delta \theta = 100^{\circ} C$.

$\Delta V = 1000 \times (182 \times 10^{-6} - 30 \times 10^{-6}) \times 100$

$\Delta V = 1000 \times (152 \times 10^{-6}) \times 100 = 15.2 \text{ cc}$

Thus, the amount of mercury that overflows is $15.2 \text{ cc}$.

(A) The change in volume $(\Delta V)$ due to an increase in temperature $(\Delta T)$ is given by:
$\Delta V = \alpha_V V \Delta T$ ... $(i)$
where $\alpha_V$ is the coefficient of volume expansion and $V$ is the initial volume of the gas.
For an ideal gas,the equation of state is:
$pV = nRT$ ... $(ii)$
At constant pressure $p$,differentiating the ideal gas equation gives:
$p \Delta V = nR \Delta T$
$\Delta V = \frac{nR \Delta T}{p}$ ... $(iii)$
Comparing equations $(i)$ and $(iii)$:
$\alpha_V V \Delta T = \frac{nR \Delta T}{p}$
$\alpha_V = \frac{nR}{pV}$
Substituting $pV = nRT$ from equation $(ii)$:
$\alpha_V = \frac{nR}{nRT} = \frac{1}{T}$
Thus,$\alpha_V = \frac{1}{T}$.

(B) The volume of mercury that overflows is equal to the difference in the volume expansion of mercury and the glass flask.
$
\Delta V = V_0 [\gamma_m - \gamma_g] \Delta \theta
$
Since $\gamma_g = 3\alpha_g$,we have:
$
\Delta V = V_0 [\gamma_m - 3\alpha_g] \Delta \theta
$
Given:
$V_0 = 1 \ L = 1000 \ cc$
$\gamma_m = 1.82 \times 10^{-4} /{ }^{\circ} C = 182 \times 10^{-6} /{ }^{\circ} C$
$\alpha_g = 0.1 \times 10^{-4} /{ }^{\circ} C = 10 \times 10^{-6} /{ }^{\circ} C$
$\Delta \theta = 100^{\circ} C - 0^{\circ} C = 100^{\circ} C$
Substituting the values:
$
\Delta V = 1000 \times [182 \times 10^{-6} - 3(10 \times 10^{-6})] \times 100
$
$
\Delta V = 1000 \times [182 \times 10^{-6} - 30 \times 10^{-6}] \times 100
$
$
\Delta V = 1000 \times [152 \times 10^{-6}] \times 100 = 15.2 \ cc
$

(D) The coefficient of real expansion of a liquid $(\gamma_r)$ is constant and is given by the sum of the coefficient of apparent expansion $(\gamma_a)$ and the coefficient of volume expansion of the container $(\gamma_v = 3\alpha)$.
For the copper vessel: $\gamma_r = \gamma_{a1} + 3\alpha_{cu}$.
For the steel vessel: $\gamma_r = \gamma_{a2} + 3\alpha_{st}$.
Equating the two expressions for $\gamma_r$: $\gamma_{a1} + 3\alpha_{cu} = \gamma_{a2} + 3\alpha_{st}$.
Given: $\gamma_{a1} = 6 \times 10^{-6} /{ }^{\circ} C$, $\gamma_{a2} = 24 \times 10^{-6} /{ }^{\circ} C$, and $\alpha_{cu} = 18 \times 10^{-6} /{ }^{\circ} C$.
Substituting the values: $6 \times 10^{-6} + 3(18 \times 10^{-6}) = 24 \times 10^{-6} + 3\alpha_{st}$.
$6 \times 10^{-6} + 54 \times 10^{-6} = 24 \times 10^{-6} + 3\alpha_{st}$.
$60 \times 10^{-6} = 24 \times 10^{-6} + 3\alpha_{st}$.
$3\alpha_{st} = 36 \times 10^{-6}$.
$\alpha_{st} = 12 \times 10^{-6} /{ }^{\circ} C$.

(C) The coefficient of real expansion $(\gamma_r)$ of a liquid is the sum of the coefficient of apparent expansion $(\gamma_a)$ and the coefficient of volume expansion of the container $(\gamma_g)$.
Mathematically,$\gamma_r = \gamma_a + \gamma_g$.
Since the coefficient of volume expansion $(\gamma_g)$ is three times the coefficient of linear expansion $(\alpha_g)$ for an isotropic solid,we have $\gamma_g = 3 \alpha_g$.
Substituting this into the first equation,we get $\gamma_r = \gamma_a + 3 \alpha_g$.

(B) Density at $0^{\circ} C, \rho_0 = 1.0127 \ g/cm^3$
Density at $100^{\circ} C, \rho_{100} = 1 \ g/cm^3$
Coefficient of real expansion of liquid,$\gamma_{\text{real}} = \frac{\rho_0 - \rho_{100}}{\rho_{100} \times \Delta t}$
$\gamma_{\text{real}} = \frac{1.0127 - 1}{1 \times 100} = \frac{0.0127}{100} = 1.27 \times 10^{-4} /^{\circ} C$
Coefficient of volume expansion of glass,$\gamma_g = 3 \alpha = 3 \times 9 \times 10^{-6} = 27 \times 10^{-6} = 0.27 \times 10^{-4} /^{\circ} C$
Coefficient of apparent expansion,$\gamma_{\text{app}} = \gamma_{\text{real}} - \gamma_g$
$\gamma_{\text{app}} = 1.27 \times 10^{-4} - 0.27 \times 10^{-4} = 1 \times 10^{-4} /^{\circ} C$
Using the formula for apparent expansion: $\gamma_{\text{app}} = \frac{\Delta m}{m_2 \Delta t}$,where $\Delta m$ is the mass expelled and $m_2$ is the remaining mass.
Alternatively,$\Delta m = m_1 \times \gamma_{\text{app}} \times \Delta t$ (approximate for small expansion) or more accurately:
$V_0 = \frac{m_1}{\rho_0}$. At $100^{\circ} C$,$V_{100} = V_0(1 + \gamma_g \Delta t) = \frac{m_1}{\rho_0}(1 + \gamma_g \Delta t)$.
Mass remaining $m_2 = V_{100} \times \rho_{100} = \frac{m_1 \rho_{100}}{\rho_0}(1 + \gamma_g \Delta t) = \frac{300 \times 1}{1.0127}(1 + 27 \times 10^{-6} \times 100) \approx 296.24 \ g$.
Mass expelled $= m_1 - m_2 = 300 - \frac{300}{1.01} = \frac{3}{1.01} \ g$.

(D) The coefficient of real expansion of a liquid $(\gamma_{\text{real}})$ is constant for a given liquid. The relationship between real expansion, apparent expansion $(\gamma_{\text{app}})$, and the volume expansion of the vessel $(\gamma_{\text{vessel}})$ is given by: $\gamma_{\text{real}} = \gamma_{\text{app}} + \gamma_{\text{vessel}}$.
For vessel $A$, the coefficient of volume expansion is $\gamma_A = 3\alpha$. Thus, $\gamma_{\text{real}} = \gamma_1 + 3\alpha$.
For vessel $B$, let the coefficient of linear expansion be $\alpha_B$. Then $\gamma_B = 3\alpha_B$. Thus, $\gamma_{\text{real}} = \gamma_2 + 3\alpha_B$.
Since $\gamma_{\text{real}}$ is the same in both cases, we equate them: $\gamma_1 + 3\alpha = \gamma_2 + 3\alpha_B$.
Solving for $\alpha_B$: $3\alpha_B = \gamma_1 - \gamma_2 + 3\alpha$.
Therefore, $\alpha_B = \frac{\gamma_1 - \gamma_2}{3} + \alpha$.

(B) Let $\gamma_L$ be the real coefficient of volume expansion of the liquid.
Let $\gamma_C$ and $\gamma_S$ be the coefficients of volume expansion of copper and silver,respectively.
We know that the coefficient of volume expansion $\gamma = 3 \times \text{linear coefficient of expansion } \alpha$.
Given $\gamma_C = 3A$.
The apparent expansion coefficient is given by $\gamma_{app} = \gamma_L - \gamma_{vessel}$.
For copper: $C = \gamma_L - 3A \implies \gamma_L = C + 3A$.
For silver: $S = \gamma_L - \gamma_S \implies \gamma_S = \gamma_L - S$.
Substituting $\gamma_L = C + 3A$ into the equation for silver: $\gamma_S = C + 3A - S$.
Since $\gamma_S = 3 \alpha_S$,where $\alpha_S$ is the linear coefficient of expansion of silver:
$3 \alpha_S = C + 3A - S \implies \alpha_S = \frac{C + 3A - S}{3}$.