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)<br /><br />a) To find the average velocity of the molecules in a sample of oxygen at $100^{\circ }C$, we can use the root mean square (RMS) velocity formula:<br /><br />$v_{rms} = \sqrt{\frac{3RT}{M}}$<br /><br />where $R$ is the universal gas constant ($8.314 \, J/(mol \cdot K)$), $T$ is the temperature in Kelvin, and$ is the molar mass of oxygen ($32.00 \, g/mol$).<br /><br />First, convert the temperature to Kelvin:<br /><br />$^{\circ }C + 273.15 = 373.15 \, K$<br /><br />Now, plug in the values into the formula:<br /><br />$v_{rms} = \sqrt{\frac{3 \times 8.314 \times 373.15}{32.00 \times 10^{-3}}} \approx 482.5 \, m/s$<br /><br />b) To find the heat flux through the wall, we can use the formula:<br /><br />$q = \frac{k(T_1 - T_2)}{d}$<br /><br />where $k$ is the thermal conductivity of the brick ($0.6 \, W/(m \cdot K $1$ and $T_2$ are the temperatures on the inside and outside of the wall ($16^{\circ }C$ and $6^{\circ }C$), and $d$ is the thickness of the wall ($0.3 \, m$).<br /><br />First, convert the temperatures to Kelvin:<br /><br />$T_1 = 16^{\circ }C + 273.15 = 289.15 \, K$<br /><br />$T_2 = 6^{\circ }C + 273.15 = 279 K$<br /><br />Now, plug in the values into the formula:<br /><br />$q = \frac{0.6 \times (289.15 - 279.15)}{0.3} = 20 \, W/m^2 find the total rate of heat loss through the wall, multiply the heat flux by the area of the wall:<br /><br />$Q = q \times A = 20 \, W/m^2 \times 7 \, m \times 6 \, m = 840 \, W$<br /><br />c) i) To find the initial volume of the gas, we can use the ideal gas law:<br /><br />$PV = nRT$<br /><br />where $P$ is the pressure ($2.0 \, atm$), $V$ is the volume, $n$ is the number of moles (1 mole), $R$ is the universal gas constant ($0.0821 \, L \cdot atm/(mol \cdot K)$), and $T$ is the temperature in Kelvin ($30^{\circ }C +.15 = 303.15 \, K$).<br /><br />Rearranging the equation to solve for $V$:<br /><br />$V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 303.15}{2.0} \approx 12.43 \, L$<br /><br />ii) To find how much the lid rises when the gas is heated to $50^{\circ }C$, we can use the ideal gas law again. The final volume $V_2$ can be found using the formula:<br /><br />$\frac{V_1}{T_1} = \frac{V_2}{T_2}$<br /><br />where $T_2$ is the final temperature in Kelvin ($50^{\circ }C + 273.15 = 323.15 \, K$).<br /><br />$V_2 = V_1 \times \frac{T_2}{T_1} = 12.43 \, L \times \frac{323.15}{303.15} \approx 13.23 \, L$<br /><br />The lid rises by $13.23 \, L - 12.43 \, L = 0.8 \, L$.<br /><br />iii) To find the final pressure in Pa when the gas is heated to $100^{\circ }C$, we can use the ideal gas law again. The final pressure $P_2$ can be found using the formula:<br /><br />$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$<br /><br />where $T_2$ is the Kelvin (${\circ }C + 273.15 = 373.15 \, K$).<br /><br />$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