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Here's the solution to your question: a) State TWO assumptions used to derive this equation. Here are two assumptions used in the kinetic theory of gases: • The gas consists of a very large number of identical molecules that are in random, continuous motion. • The volume occupied by the gas molecules themselves is negligible compared to the total volume of the gas. • Collisions between molecules and with the container walls are perfectly elastic, and the time duration of collisions is negligible. • There are no intermolecular forces between gas molecules, except during collisions. • The average kinetic energy of the gas molecules is directly proportional to the absolute temperature of the gas. b) Use the above equation to show that c^2 is related to the absolute temperature of the gas by; c^2 = (3RT)/(M) Step 1: Start with the given kinetic theory equation for pressure. P = (1)/(3) c^2 Step 2: Express density () in terms of mass (m), volume (V), and molar mass (M). Density is mass per unit volume, = (m)/(V). The total mass of the gas (m) can also be expressed as the number of moles (n) multiplied by the molar mass (M), so m = nM. Therefore, = (nM)/(V). Step 3: Substitute the expression for density into the pressure equation. P = (1)/(3) ((nM)/(V)) c^2 Step 4: Rearrange the equation to isolate PV. PV = (1)/(3) n M c^2 Step 5: Use the Ideal Gas Law, PV = nRT. Equate the expression for PV from Step 4 with the Ideal Gas Law: nRT = (1)/(3) n M c^2 Step 6: Cancel n from both sides and solve for c^2. RT = (1)/(3) M c^2 Multiply both sides by 3 and divide by M: c^2 = (3RT)/(M) This shows the relationship between the mean-square speed (c^2) and the absolute temperature (T). c) Determine the root-mean-square speed of the molecules if the gas is heated to a temperature of 150 °C. Step 1: Convert the given temperatures from Celsius to Kelvin. Initial temperature, T_1 = 30.0 °C + 273.15 = 303.15 K Final temperature, T_2 = 150 °C + 273.15 = 423.15 K Step 2: Relate the root-mean-square speed to temperature. From part (b), we know that c^2 = (3RT)/(M). This implies that the mean-square speed (c^2) is directly proportional to the absolute temperature (T). Therefore, the root-mean-square speed (c_rms = sqrt(c^2)) is proportional to the square root of the absolute temperature: c_rms sqrt(T) So, we can write the ratio: c_rms,1sqrt(T_1) = c_rms,2sqrt(T_2) Step 3: Substitute the known values and solve for the final root-mean-square speed (c_rms,2). Given initial rms speed, c_rms,1 = 850 m s^-1. c_rms,2 = c_rms,1 sqrt((T_2)/(T_1)) c_rms,2 = 850 m s^-1 sqrt(423.15 K)303.15 K c_rms,2 = 850 m s^-1 sqrt(1.395844) c_rms,2 = 850 m s^-1 × 1.181458 c_rms,2 ≈ 1004.24 m s^-1 Step 4: Round the answer to an appropriate number of significant figures (3 significant figures, based on the input values). c_rms,2 ≈ 1000 m s^-1 The root-mean-square speed of the molecules if the gas is heated to 150 °C is 1,00 × 10^3 m s^-1. Send me the next one 📸