This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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6. Six grams of oxygen gas at a temperature of 30^ have a volume of 0.00142 m^3. Calculate: a) The pressure in the gas if the gas constant for oxygen is 280 J/kg K. Step 1: Identify the given values and convert units to SI. Mass, m = 6 g = 0.006 kg Temperature, T_1 = 30^ Volume, V_1 = 0.00142 m^3 Gas constant for oxygen, R = 280 J/kg K Convert the temperature to Kelvin: T_1 = 30^ + 273.15 = 303.15 K We will use the Ideal Gas Law: P_1V_1 = mRT_1. Step 2: Rearrange the Ideal Gas Law to solve for the pressure P_1. P_1 = (mRT_1)/(V_1) Step 3: Substitute the values and calculate P_1. P_1 = (0.006 kg) × (280 J/kg K) × (303.15 K)0.00142 m^3 P_1 = (509.292)/(0.00142) Pa P_1 ≈ 358656.338 Pa Convert to kPa and round to one decimal place: P_1 ≈ 358.7 kPa P_1 ≈ 358.7 kPa b) The final volume of the gas if the pressure is tripled and the temperature increases to 130^. Step 1: Identify the initial and final states and convert temperatures to Kelvin. Initial pressure, P_1 = 358.656 kPa (using the unrounded value from part a) Initial volume, V_1 = 0.00142 m^3 Initial temperature, T_1 = 303.15 K Final pressure, P_2 = 3 × P_1 = 3 × 358.656 kPa = 1075.968 kPa Final temperature, T_2 = 130^ Convert the final temperature to Kelvin: T_2 = 130^ + 273.15 = 403.15 K We will use the Combined Gas Law: (P_1V_1)/(T_1) = (P_2V_2)/(T_2). Step 2: Rearrange the Combined Gas Law to solve for the final volume V_2. V_2 = (P_1V_1T_2)/(P_2T_1) Step 3: Substitute the values and calculate V_2. V_2 = (358.656 kPa) × (0.00142 m^3) × (403.15 K)(1075.968 kPa) × (303.15 K) V_2 = (204.887)/(326.066) m^3 V_2 ≈ 0.0005999 m^3 Rounding to five decimal places: V_2 ≈ 0.00060 m^3 Drop the next question.