Вопрос
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)
Решения
4.1
(261 Голоса)
Полина
мастер · Репетитор 5 лет
Ответ
a) To determine the volume of the spilled water, we need to calculate the change in volume of the aluminum container due to heating.The change in volume of the aluminum container can be calculated using the formula:ΔV = V * β * ΔTWhere:V = Initial volume of the container (200 cm³)β = Coefficient of volume expansion of aluminum (6.3 x 10⁻⁴ °C⁻¹)ΔT = Change in temperature (90°C - 0°C = 90°C)Substituting the values, we get:ΔV = 200 cm³ * 6.3 x 10⁻⁴ °C⁻¹ * 90°CΔV = 11.22 cm³Since 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.b) To determine the specific latent heat of fusion of ice, we can use the formula:Q = mLWhere:Q = Heat energy supplied (Power * Time)m = Mass of ice melted (34 g)L = Specific latent heat of fusion of iceThe heat energy supplied can be calculated as:Q = Power * TimeQ = 100 W * 4 min * 60 s/minQ = 24000 JSubstituting the values in the formula, we get:24000 J = 34 g * LL = 24000 J / 34 gL = 705.88 J/gTherefore, the specific latent heat of fusion of ice is 705.88 J/g.c) The work done in compressing an ideal gas can be derived from the first law of thermodynamics:W = ΔU + PΔVFor 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:W = PΔVUsing the ideal gas law, we can express the pressure (P) as:P = nRT/VSubstituting this expression for pressure into the work equation, we get:W = nRT/V * ΔVW = nRT * (V₂ - V₁) / V₁W = nRT * ln(V₂/V₁)Therefore, the work done in compressing an ideal gas is W = nRT * ln(V₂/V₁).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:P₁V₁ = P₂V₂Where:P₁ = Initial pressureV₁ = Initial volumeP₂ = Final pressureV₂ = Final volumeii. To determine the volume of the gas at standard temperature and pressure (STP), we can use Boyle's law.Given:Initial temperature = 20°CInitial pressure = 700 mmHgInitial volume = 120 cm³We need to find the final volume (V₂) at STP, which is 0°C and 760 mmHg.Using Boyle's law, we can set up the equation:P₁V₁ = P₂V₂Substituting the given values, we get:700 mmHg * 120 cm³ = 760 mmHg * V₂Solving for V₂, we get:V₂ = (700 mmHg * 120 cm³) / 760 mmHgV₂ = 105.26 cm³Therefore, the volume of the gas at STP is 105.26 cm³.