Assuming ideal behavior, which of these gas samples has the greatest volume at stp?

Assuming ideal behavior, which of these gas samples has the greatest volume at stp? assuming ideal behavior, which of these gas samples has the greatest volume at stp? 1 g of o2 1 g of h2 1 g of ar

Mole of O₂     =  1 g / 32 g.mol⁻¹  =  0.03125 mol                   Mole of H₂     =  1 g / 2.016 g.mol⁻¹  =  0.0496 mol                   Mole of Ar     =  1 g / 39.94 g.mol⁻¹  =  0.025 mol Solution:As Ideal Gas Equation is given as,                                    P V  =  n R TSolving for V,                                      V  =  n R T / P Solving for O₂;                             V  = (0.03125 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 273 K) ÷ 1 atm               V  =  0.699 L Solving for H₂;                             V  = (0.0496 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 273 K) ÷ 1 atm               V  =  1.11 L Solving for Ar;                             V  = (0.025 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 273 K) ÷ 1 atm               V  =  0.55 L Result:            The correct answer is H₂. 1 g of H₂ occupies more volume as compare to 1 g of O₂ and 1 g of Ar.

2) Kinetic energy is given by the formula 1/2mv^2, where m is the mass and v is the velocity. All the gases have the same kinetic energy, mass and velocity are inversely proportional. Therefore, the most massive molecules will have the lowest average speed. H2 is the fastest, Xe is the slowest.Xe, Cl2, O2, H2, in the order of increasing speed.  3) Two separate gases will have the same kinetic energy at the same temperature, but not the same average speeds. To have the same average speed the two gases will not be at the same temperature.

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Option (a) is the correct answer. Explanation: At low pressure and high temperature there exists no force of attraction or repulsion between the molecules of a gas. Hence, gases behave ideally at these conditions.
Whereas at low temperature there occurs a decrease in kinetic energy of gas molecules and high pressure causes the molecules to come closer to each other.   As a result, there exists force of attraction between the molecules at low temperature and high pressure and under these conditions gases are known as real gases. Thus, we can conclude that the ideal gas law tends to become inaccurate when the pressure is raised and the temperature is lowered.

The correct option is D. Explanation: To calculate the larges number of partivles of gas, we will use Avogadro’s Law which says that volume of the gas is directly related to the number of moles of gas at constant pressure and temperature. Mathematically,       ….(1) The gas which has the largest number of moles, will have the largest number of particles. We are provided to use STP conditions, which says that 1 mole of a gas occupies 22.4 L of volume. Initial conditions: A. 0.520 L of Putting all the values, in equation 1, we get B. 0.10 L of Xe Putting all the values, in equation 1, we get C. 7.0 L of Putting all the values, in equation 1, we get D. 12.0 L of Ne Putting all the values, in equation 1, we get Hence, from the above calculations, we see that option D has the largest amount of moles and hence, will have the largest amount of particles.

Answer 6

Answer : The correct statement is (C). Explanation : Statement A : This statement is incorrect because the temperature of the gas sample is directly related to the average kinetic energy of the gas particles. Statement B : This statement is incorrect because at STP, equal volume of all gases have the same number of molecules. Statement D : This statement is incorrect because the ideal gas works well when the inter-molecular force of attraction between the gas molecules are negligible. Statement C : This statement is correct because when molecules collides with the wall of container, they exert small force on the wall of the container. The pressure exerted by the gas is due to the sum of all the collision forces. That means more the particles hit the walls of container, the higher the pressure.

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Answer 7

Answer; N2O (Nitrogen (i) oxide. Explanation; The decomposition would be; 2N2O(g) = 2N2 (g)+ O2 (g) Which means the ratio of; N2O :N2 :O2 = 2:2:1 The volume of colorless gas is 2.5 liters, N2 is 2.5 liters and oxygen is 1.25 liters; which also gives us a ratio of; 2: 2: 1. Alternatively; here is twice as much N2 as O2. therefore; the formula would be N2O
(nitrogen dioxide) to make 1 volume of N2 for every 2 volumes of O2.

Explanation: The given data is as follows.         Mass of = 1.15 g         Mass of = 1.55 g Therefore, moles of oxygen present will be as follows.              No. of moles of =                               =                               = 0.035 mol              No. of moles of =                               =                               = 0.055 mol Hence, total no. of moles = moles of + moles of                               = (0.035 + 0.055) mol                               = 0.09 mol Now, it is known that at STP volume is 22.4 L/mol. Hence, volume of the gas sample at STP for 0.09 moles will be as follows.                                           = 2.01 L Thus, we can conclude that volume of the given gas sample is 2.01 L.

Explanation: The given data is as follows.         Mass of = 1.15 g         Mass of = 1.55 g Therefore, moles of oxygen present will be as follows.              No. of moles of =                               =                               = 0.035 mol              No. of moles of =                               =                               = 0.055 mol Hence, total no. of moles = moles of + moles of                               = (0.035 + 0.055) mol                               = 0.09 mol Now, it is known that at STP volume is 22.4 L/mol. Hence, volume of the gas sample at STP for 0.09 moles will be as follows.                                           = 2.01 L Thus, we can conclude that volume of the given gas sample is 2.01 L.

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Molar mass of the unknown gas: 146 g /mol Explanation: At STP, any “ideal” gas is contained in a volume of 22.4L. This situation, for 1 mol of gas. In this case we have: 0.5 g / 4 g/mol = 0.125 moles of He These amount of He occupies 0.125 mol . 22.4L / 1 mol = 2.8 L so the double of volume will be occupied by the unkown gas. 22.4 L / 1 mol = (2.8 L . 2) / x moles 22.4L / 1 mol = 5.6 L / x mol x mol = 5.6L / 22.4L → 0.25 moles To determine the molar mass of the unknown gas → 36.5 g / 0.25 mol = 146 g /mol

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