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P\(_1\)V\(_1\) = P\(_2\)V\(_2\) When the bubble comes to the surface and pressure drops 3 times, the volume increases 3 times. V\(_2\) = 3 x 2 = 6 cm\(^3\)

To find the ratio of the final volume of the gas to its initial volume, we use the ideal gas law: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) Rearranging gives: \(\frac{V_2}{V_1} = \frac{P_1 T_2}{P_2 T_1}\) Substituting the values: - \( P_1 = 200 \, N/m^2 \) - \( P_2 = 100 \, N/m^2 \) - \( T_1 = 400.15 \, K \) - \( T_2 = 200.15 \, K \) We have: \(\frac{V_2}{V_1} = \frac{200 \times 200.15}{100 \times 400.15}\) Calculating the ratio: \(\frac{V_2}{V_1} = \frac{40030}{40015} \approx 1.0:1\) Thus, the ratio of the final volume of the gas to its initial volume is: 1.0:1

To calculate the apparent cubic expansion of the mercury, we use the formula for apparent expansion: \(\text{Apparent Expansion} = V_0 \cdot \left( \beta_m - \beta_g \right) \cdot \Delta T\) Where: - \( V_0 = 100 \, \text{cm}^3 \) (initial volume of mercury) - \( \beta_m = 1.82 \times 10^{-4} \, \text{K}^{-1} \) (real cubic expansivity of mercury) - \( \beta_g = 2.4 \times 10^{-5} \, \text{K}^{-1} \) (cubic expansivity of glass) - \( \Delta T = 100^\circ C - 10^\circ C = 90 \, \text{K} \) (change in temperature) First, we calculate \( \beta_m - \beta_g \): \(\beta_m - \beta_g = 1.82 \times 10^{-4} - 2.4 \times 10^{-5} = 1.58 \times 10^{-4} \, \text{K}^{-1}\)) Now substitute the values into the apparent expansion formula: \(\text{Apparent Expansion} = 100 \, \text{cm}^3 \cdot (1.58 \times 10^{-4}) \cdot 90\) Calculating this gives: \(= 100 \cdot 1.58 \times 10^{-4} \cdot 90\) \(= 100 \cdot 1.422 \times 10^{-2}\) \(= 1.422 \, \text{cm}^3\) Thus, the apparent cubic expansion of the mercury is approximately: \(\text{Answer: } 1.42 \, \text{cm}^3\)

No expanded working has been attached for this question yet. The verified answer is Option C.

The formula of specific latent heat of fusion is given by: Q = mL : 20 X 336 = 6720 or 6.72 X 10\(^3\)

a) The boiling pointer of a liquid is the temperature at which its s.v.p. is equal to the external atmospheric pressure OR The temperature at which a (pure) liquid boils when the atmospheric pressure on it is 760 mm Hg. OR The temperature at which the liquid changes from the liquid to the gaseous state without change in temperature when the atmospheric pressure is normal (760mm Hg). Method: Some mercury is introduced into J - tube and the entrapped air removed. Some liquid is introduced into the shorter length of the J - tube above the mercury. The beaker and its contents are heated and the water stirred continuously. At a stage when the levels of mercury in the two limbs of the J-tube are the same the temperature is read. At this temperature the s.v.p of the liquid is equal to the atmospheric pressure, and this temperature is the boiling point of the liquid. The process is repeated with the beaker and its contents cooling. The temperature at which the levels of mercury in the two limbs are equal is read and the mean of the two taken. (c) Pressure on the liquid, impurities in the liquid. (d) The extra heat supplied during boiling is used in breaking the intermolecular attractions and in mabling the steam expand or enabling the molecules move further apart against the atmospheric pressure. Hence, there is no change in temperature during the change of state.

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option E.

No expanded working has been attached for this question yet. The verified answer is Option E.

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option E.

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option B.

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option A.