Question Five a) A 200cm^3 aluminum container is filled with water at 0^circ C and then heated to 90^circ C. If the coefficient at linear expansion of aluminum is 24times 10^-6C^-1 and the coefficient of volume expansion of water is 6.3times 10^-4C^-1 Determine the volume of the spilled water. (5mks) b) An immersion heater rated 100w is placed in melting ice for 4 min. if the mass of ice melted was 34g, determine the specific latent heat of fusion of ice. (5mks) c) Show that the work done in compressing an ideal gas W=nRTInV_(2)/V_(1). (5mks) d) i. State Boyles law (1mk) ii. The volume of a gas at 20^circ C and 700mm Hg pressure is 120cm^3 Determine its volume at stp. (4mks)

a) To determine the volume of the spilled water, we need to calculate the change in volume of the aluminum container due to heating.<br /><br />The change in volume of the aluminum container can be calculated using the formula:<br /><br />ΔV = V * β * ΔT<br /><br />Where:<br />V = Initial volume of the container (200 cm³)<br />β = Coefficient of volume expansion of aluminum (6.3 x 10⁻⁴ °C⁻¹)<br />ΔT = Change in temperature (90°C - 0°C = 90°C)<br /><br />Substituting the values, we get:<br /><br />ΔV = 200 cm³ * 6.3 x 10⁻⁴ °C⁻¹ * 90°C<br />ΔV = 11.22 cm³<br /><br />Since the volume of water spilled is equal to the change in volume of the aluminum container, the volume of the spilled water is 11.22 cm³.<br /><br />b) To determine the specific latent heat of fusion of ice, we can use the formula:<br /><br />Q = mL<br /><br />Where:<br />Q = Heat energy supplied (Power * Time)<br />m = Mass of ice melted (34 g)<br />L = Specific latent heat of fusion of ice<br /><br />The heat energy supplied can be calculated as:<br /><br />Q = Power * Time<br />Q = 100 W * 4 min * 60 s/min<br />Q = 24000 J<br /><br />Substituting the values in the formula, we get:<br /><br />24000 J = 34 g * L<br />L = 24000 J / 34 g<br />L = 705.88 J/g<br /><br />Therefore, the specific latent heat of fusion of ice is 705.88 J/g.<br /><br />c) The work done in compressing an ideal gas can be derived from the first law of thermodynamics:<br /><br />W = ΔU + PΔV<br /><br />For an ideal gas, the change in internal energy (ΔU) is zero, as there is no change in temperature. Therefore, the work done is equal to the pressure-volume work:<br /><br />W = PΔV<br /><br />Using the ideal gas law, we can express the pressure (P) as:<br /><br />P = nRT/V<br /><br />Substituting this expression for pressure into the work equation, we get:<br /><br />W = nRT/V * ΔV<br />W = nRT * (V₂ - V₁) / V₁<br />W = nRT * ln(V₂/V₁)<br /><br />Therefore, the work done in compressing an ideal gas is W = nRT * ln(V₂/V₁).<br /><br />d) i. Boyle's law states that the pressure of a gas is inversely proportional to its volume, provided the temperature remains constant. Mathematically, it can be expressed as:<br /><br />P₁V₁ = P₂V₂<br /><br />Where:<br />P₁ = Initial pressure<br />V₁ = Initial volume<br />P₂ = Final pressure<br />V₂ = Final volume<br /><br />ii. To determine the volume of the gas at standard temperature and pressure (STP), we can use Boyle's law.<br /><br />Given:<br />Initial temperature = 20°C<br />Initial pressure = 700 mmHg<br />Initial volume = 120 cm³<br /><br />We need to find the final volume (V₂) at STP, which is 0°C and 760 mmHg.<br /><br />Using Boyle's law, we can set up the equation:<br /><br />P₁V₁ = P₂V₂<br /><br />Substituting the given values, we get:<br /><br />700 mmHg * 120 cm³ = 760 mmHg * V₂<br /><br />Solving for V₂, we get:<br /><br />V₂ = (700 mmHg * 120 cm³) / 760 mmHg<br />V₂ = 105.26 cm³<br /><br />Therefore, the volume of the gas at STP is 105.26 cm³.