QUESTION THREF (20 MARKS) a) What is the average velocity of the molecules in a sample of oxygen at 100^circ C The mass of an oxygen molecule is 5.3times 10^-26 Kg. (4 marks) b) The wall of a house, 7 m wide and 6 m high is made from 0.3 m thick brick with k=0.6(W)/(mK) The surface temperature on the inside of the wall is 16^circ C and that on the outside is 6^circ C. Find the heat flux through the wall and the total rate of heat loss through wall. (6 marks) c) One mole of nitrogen gas at 30^circ C is placed in a sealed container with a moveable lid with an area of 100cm^2. The pressure in the container is 2.0 atm . (this is the initial state) i) Calculate the initial volume of the gas. (4 marks) ii) If the gas is heated to 50^circ C, how much does the lid rise? (3 marks) Page 2 of 3 iii) If the lid is held in place and the gas is heated to 100^circ C find the final pressure in Pa. (3 marks) QUESTION FOUR (20 MARKS) a) State: i) The first law of thermodynamics in terms of internal energy and state the significance of the law. (3 marks) ii) A gas is compressed at a constant pressure of 0.5 atm from 5.0 litres to 2.0 lit- res, losing 500 J of heat in the process. Find the change in internal energy. (5 marks) b) A 200 g ice initially at -20^circ C is heated to become water vapour at 100^circ C. What amount of heat energy is used in this process? (7marks) c) The sun radiates energy at the rate of 6.5times 10^7W/m^2 from its surface. Assuming that the sun radiates as a blackbody (which is approximately true), find its surface temperature? The emissivity of a blackbody is e =1. (5 marks)

QUESTION THREE (20 MARKS)a) To find the average velocity of the molecules in a sample of oxygen at , we can use the root mean square (RMS) velocity formula: where is the universal gas constant ( ), is the temperature in Kelvin, and 32.00 \, g/mol ^{\circ }C + 273.15 = 373.15 \, K v_{rms} = \sqrt{\frac{3 \times 8.314 \times 373.15}{32.00 \times 10^{-3}}} \approx 482.5 \, m/s q = \frac{k(T_1 - T_2)}{d} k 0.6 \, W/(m \cdot K and are the temperatures on the inside and outside of the wall ( and ), and is the thickness of the wall ( ).First, convert the temperatures to Kelvin: Now, plug in the values into the formula: Q = q \times A = 20 \, W/m^2 \times 7 \, m \times 6 \, m = 840 \, W PV = nRT P 2.0 \, atm V n R 0.0821 \, L \cdot atm/(mol \cdot K) T 30^{\circ }C +.15 = 303.15 \, K V V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 303.15}{2.0} \approx 12.43 \, L 50^{\circ }C V_2 \frac{V_1}{T_1} = \frac{V_2}{T_2} T_2 50^{\circ }C + 273.15 = 323.15 \, K V_2 = V_1 \times \frac{T_2}{T_1} = 12.43 \, L \times \frac{323.15}{303.15} \approx 13.23 \, L 13.23 \, L - 12.43 \, L = 0.8 \, L 100^{\circ }C P_2 \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} T_2 {\circ }C + 273.15 = 373.15 \, K P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 2.0 \, atm \times \frac{12.43 \, L}{13.23 \, L} \times \frac{373.15