You're on a roll — Step 1: Calculate the change in temperature ($\Delta T$). $$\Delta T = T_{final} - T_{initial}$$ $$\Delta T = 70 ^\circ\text{C} - 20 ^\circ\text{C}$$ $$\Delta T = 50 ^\circ\text{C}$$ Step 2: Calculate the change in length ($\Delta L$) using the linear thermal expansion formula. $$\Delta L = L_0 \alpha \Delta T$$ Where: $L_0 = 100 \text{ cm}$ (original length) $\alpha = 0.000019 \text{ per } ^\circ\text{C}$ (linear expansivity) $\Delta T = 50 ^\circ\text{C}$ (change in temperature) Substitute the values: $$\Delta L = (100 \text{ cm}) \times (0.000019 \text{ per } ^\circ\text{C}) \times (50 ^\circ\text{C})$$ $$\Delta L = 0.000019 \times 5000 \text{ cm}$$ $$\Delta L = 0.095 \text{ cm}$$ Step 3: Calculate the final length ($L$). $$L = L_0 + \Delta L$$ $$L = 100 \text{ cm} + 0.095 \text{ cm}$$ $$L = 100.095 \text{ cm}$$ The correct option is A. $$\boxed{\text{A. } 100.095 \text{ cm}}$$ Absolute zero temperature is defined as the lowest possible temperature, where all molecular motion ceases. This corresponds to $0 \text{ K}$ or approximately $-273 ^\circ\text{C}$. The correct option is B. $$\boxed{\text{B. } -273 ^\circ\text{C}}$$ Boyle's Law states that for a fixed mass of gas at constant temperature, the pressure of the gas is inversely proportional to its volume. This can be expressed as $P \propto \frac{1}{V}$ or $PV = \text{constant}$. The correct option is B. $$\boxed{\text{B. Inversely proportional to volume}}$$ The temperature at which water vapour in the air just saturates is called the dew point. At this temperature, the air can no longer hold all of the water vapour, and condensation begins to occur. The correct option is C. $$\boxed{\text{C. Dew point}}$$ Step 1: Convert temperatures from Celsius to Kelvin. Charles's Law requires temperatures to be in Kelvin. $$T_1 = 27 ^\circ\text{C} + 273 = 300 \text{ K}$$ $$T_2 = 127 ^\circ\text{C} + 273 = 400 \text{ K}$$ Step 2: Apply Charles's Law. Charles's Law states that for a fixed mass of gas at constant pressure, the volume is directly proportional to its absolute temperature: $$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$ Given: $V_1 = 200 \text{ cm}^3$ $T_1 = 300 \text{ K}$ $T_2 = 400 \text{ K}$ Step 3: Calculate the final volume ($V_2$). Rearrange the formula to solve for $V_2$: $$V_2 = V_1 \times \frac{T_2}{T_1}$$ Substitute the values: $$V_2 = 200 \text{ cm}^3 \times \frac{400 \text{ K}}{300 \text{ K}}$$ $$V_2 = 200 \text{ cm}^3 \times \frac{4}{3}$$ $$V_2 = \frac{800}{3} \text{ cm}^3$$ $$V_2 \approx 266.67 \text{ cm}^3$$ The correct option is B. $$\boxed{\text{B. } 266.7 \text{ cm}^3}$$ For an isotropic solid, the relationship between volume expansivity ($\gamma$) and linear expansivity ($\alpha$) is given by $\gamma = 3\alpha$. Given that the volume expansivity is $y$ and the linear expansivity is $a$, then: $$y = 3a$$ The correct option is C. $$\boxed{\text{C. } y = 3a}$$ Send me the next one 📸