0.55 mol of a nonlinear triatomic ideal gas with all vibrational modes frozen undergoes a process a → b shown on the following 'S diagram: T (K) اله (5 800 600 400 200 b 10 20 ind the heat, change in internal energy, and the work. 30 40 50 a 60 S (J/K)
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- On a cold day you inhale 1.0 L of air at -20 ∘C and its temperature is raised to 40∘C. Assume that you can treat the air as a diatomic ideal gas with a pressure of 1.0 atm. What is the total change in kinetic energy of the air you inhaled?Two monatomic ideal gases A and B are at the same temperature. If 1.0 g of gas A has the same internal energy as 0.10 g of gas B, what are (a) the ratio of the number of moles of each gas and (b) the ration of the atomic masses of the two gases?Consider a gas that follows the equation of state NRT aN² 2 P V - bN V2 This is called van der Waals gas (a,b > 0). For temperatures T > Tc : monotonically decreasing function. Assume T > Tc and Cy = 3NR/2. = 8a p(V) is a 27Rb Consider an insulated box with volume V₁. The box was divided by a wall. One of the compartments had volume Vo at temperature To, which was filled with van der Waals gas of amount N. The other compartment was vacuum. The system was in equilibrium. (i) The wall was removed abruptly. The gas expanded and occupied the entire box. This process is called adiabatic free expansion. The temperature in the box is now T₁ in equilibrium. What is T₁ - To? (ii) The internal energy of the gas does not change by adiabatic free expansion. Why? (iii) Instead of the abrupt removal, the wall was moved through adiabatic and quasi-static process, and the gas expanded to the entire volume of the box. The temperature in the box is now T2 in equilibrium. What is T₂? (iv) Show T₁ > T₂,…
- An ideal gas has been subjected transformations AB and BC (see pic/link) Find the change in the inner energy during the whole transformation (AB+BC), if the change in volume is 1 m3..5 moles of Neon gas have a total kinetic energy of 1.83 kJ and has a volume of 0.935 m3. Without changing the pressure the volume is change to 0.784 m3. (a)What is the average mean square velocity of a molecule at the new volume? (b) If the pressure is 1.01 x 105 Pa, What is the change in internal energy? (MW = 20.18 g/mol; u = 1.67 x 10 -27 kg)A 2.00 mol sample of an ideal diatomic gas at a pressure of 1.10 atm and temperature of 420 K undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are 720 K and 1.70 atm . Fid the change in internal energy, work done by the gas, and heat added.
- The temperature at state A is 20.0ºC, that is 293 K. During the last test, you have found the temperature at state D is 73.0 K and n = 164 moles for this monatomic ideal gas. What is the change in thermal energy for process D to B, in MJ (MegaJoules)?A sample of 2.37 moles of an ideal diatomic gas experiences a temperature increase of 65.2 K at constant volume. Find the increase in internal energy if only translational and rotational motions are possible.2.00-mol of a monatomic ideal gas goes from State A to State D via the path A→B→C→D: State A PA=11.5atm, VA=11.00L State B PB=11.5atm, VB=6.50L State C PC=20.5atm, VC=6.50L State D PD=20.5atm, VD=22.00L Assume that the external pressure is constant during each step and equals the final pressure of the gas for that step. Calculate ΔH for this process.
- The temperature of an ideal monatomic gas rises by 8.0 K. What is the change in the internal energy of 1 mol of the gas at constant volume?The temperature of 2.00 mol of an ideal monatomic gas is raised 15.0 K in an adiabatic process.What are (a) the work W done by the gas, (b) the energy transferred as heat Q, (c) the change Eint in internal energy of the gas, and (d) the change K in the average kinetic energy per atomA monatomic gas is cooled by 50 K at constant volume when 831 J of energy is removed from it. How many moles of gas are in the sample?