Answer any two numerical questions. a. A glass flask of volume 400\ cm^3 is just filled with mercury at 0^ . How much mercury will overflow when the temperature of the system rises to 80^ . b. A…

a. Mercury Overflow Due to Thermal Expansion

Given:

• Volume of flask, $V_0=400\mathrm{c}{\mathrm{m}}^3$
• Initial temperature, $T_0=0^{\circ }\mathrm{C}$
• Final temperature, $T=80^{\circ }\mathrm{C}$
• Coefficient of volume expansion of mercury, $\gamma =0.00018{}^{\circ }{\mathrm{C}}^{-1}$

Assumption: The flask is rigid, so its volume does not change with temperature. Only mercury expands.

Solution: The volume expansion of mercury is given by: $$\Delta V=V_0\gamma \Delta T$$ where $\Delta T=T-T_0=80^{\circ }\mathrm{C}-0^{\circ }\mathrm{C}=80^{\circ }\mathrm{C}$.

Substitute the values: $$\Delta V=400\mathrm{c}{\mathrm{m}}^3\times 0.00018{}^{\circ }{\mathrm{C}}^{-1}\times 80^{\circ }\mathrm{C}$$ $$\Delta V=400\times 0.0144=5.76\mathrm{c}{\mathrm{m}}^3$$

Answer: The amount of mercury that will overflow is 5.76 cm³.

---

b. Specific Heat Capacity of Aluminium

Given:

• Mass of copper calorimeter, $m_c=300\mathrm{g}$
• Mass of water, $m_w=500\mathrm{g}$
• Initial temperature of water and calorimeter, $T_{i,w}=15^{\circ }\mathrm{C}$
• Mass of aluminium block, $m_a=560\mathrm{g}$
• Initial temperature of aluminium, $T_{i,a}=100^{\circ }\mathrm{C}$
• Final equilibrium temperature, $T_f=22.5^{\circ }\mathrm{C}$
• Specific heat capacity of water, $c_w=4.18J/{\mathrm{g}}^{\circ }\mathrm{C}$
• Specific heat capacity of copper, $c_c=0.39J/{\mathrm{g}}^{\circ }\mathrm{C}$

Assumption: Heat lost by aluminium = Heat gained by water + Heat gained by calorimeter.

Solution: Using the principle of calorimetry: $$m_a{c}_a(T_{i,a}-T_f)=m_w{c}_w(T_f-T_{i,w})+m_c{c}_c(T_f-T_{i,w})$$

Rearrange to solve for $c_a$: $$c_a=\frac{m_w{c}_w(T_f-T_{i,w})+m_c{c}_c(T_f-T_{i,w})}{m_a(T_{i,a}-T_f)}$$

Substitute the values: $$c_a=\frac{500\times 4.18\times (22.5-15)+300\times 0.39\times (22.5-15)}{560\times (100-22.5)}$$ $$c_a=\frac{500\times 4.18\times 7.5+300\times 0.39\times 7.5}{560\times 77.5}$$ $$c_a=\frac{15675+877.5}{43400}$$ $$c_a=\frac{16552.5}{43400}\approx 0.381J/{\mathrm{g}}^{\circ }\mathrm{C}$$

Answer: The specific heat capacity of aluminium is 0.381 J/g°C.

---

c. Adiabatic Expansion of Dry Air

Given:

• Initial volume, $V_1=V$
• Final volume, $V_2=5V$ (expands to 5 times original volume)
• Initial conditions: NTP (Normal Temperature and Pressure)
  • $P_1=1\mathrm{atm}=1.013\times 10^5\mathrm{Pa}$
  • $T_1=273\mathrm{K}$
• Adiabatic index for dry air, $\gamma =1.4$

Assumption: Air behaves as an ideal gas undergoing reversible adiabatic expansion.

Solution: For adiabatic processes, the following relations hold: $$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$$ $$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$$

Step 1: Calculate Final Pressure ($P_2$)

$$P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }=1.013\times 10^5\times {\left(\frac{1}{5}\right)}^{1.4}$$ $$P_2=1.013\times 10^5\times (0.2{)}^{1.4}$$ $$P_2\approx 1.013\times 10^5\times 0.0856\approx 8.67\times 10^3\mathrm{Pa}$$

Step 2: Calculate Final Temperature ($T_2$)

$$T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=273\times {\left(\frac{1}{5}\right)}^{0.4}$$ $$T_2=273\times (0.2{)}^{0.4}$$ $$T_2\approx 273\times 0.66\approx 180.58\mathrm{K}$$

Answer:

• Final pressure = 8.67 × 10³ Pa
• Final temperature = 180.58 K