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# work done on the system example

What are the change in internal energy and enthalpy change of the system? (See Example.) (See Figure 7.03.2.) Figure (a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an $$F \, cos \, \theta$$ vs. $$d$$ graph. For example, if the lawn mower in [link](a) is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. In this article, we shall study to calculate the change in internal energy and change in enthalpy in a chemical reaction. We know that once the person stops pushing, friction will bring the package to rest. Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. CO reacts with O2 according to the following reaction. Suppose a 30.0-kg package on the roller belt conveyor system in Figure 7.03.2 is moving at 0.500 m/s. Given: q = + 6 kJ (Heat absorbed), W = -1.5 kJ (Work done on the surroundings). Steps. (a) The work done by the force F on this lawn mower is Fd cos θ. The enthalpy change for the following reaction is – 620 J, when 100 mL of ethylene and 100 mL of H2 react at 1 atm pressure. Does it remain in the system or move on? b) Suppose that in addition to absorption of heat by the sample, the surrounding does 2000 kJ of work on the sample. Subscribe to our Youtube Channel - https://you.tube/teachoo. Thus the total work done is the total area under the curve, a useful property to which we shall refer later. The work done by the system depends not only on the initial and final states, but also on the intermediate states—that is, on the path. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects moving together. ΔH = qp = Heat supplied at constant pressure = + 6 kJ, Ans: The change in internal energy is 4.5 kJ and enthalpy change is 6 kJ. Work done on an object transfers energy to the object. To reduce the kinetic energy of the package to zero, the work $$W_{fr}$$ by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the pushing. Calculate the work done in the following reaction when 1 mol of SO2 is oxidised at constant pressure at 5o °C. Given: q = + 4000 kJ (Heat absorbed by sample), ΔV = 0 (Volume remains the same), Work done in the process is given by   W = – Pext × ΔV = – Pext × 0 = 0, Given: q = + 4000 kJ (Heat absorbed by sample), W = + 2000 kJ (Work done by surroundings), ∴ Δ U = + 4000 kJ   + 2000 kJ = + 6000 kJ, Given: q = + 4000 kJ (Heat absorbed by sample), W = – 600 kJ (Work done on the surroundings), ∴ Δ U = + 4000 kJ   –   600 kJ = + 3400 kJ. Given:  Initial volume = V1 = 6 dm³ = 6 × 10-3 m³, Final volume = V2 = 16 dm³ = 16 × 10-3 m³, Pext = 2.026 x 105 Nm-2, ΔU = 418 J. Calculate the work done in the following reaction when 2 moles of HCl are used at constant pressure and 423 K. State whether work is on the system or by the system. Where is the energy you spend going? Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. How far does the package in Figure 7.03.2. coast after the push, assuming friction remains constant? The kinetic energy is given by $KE = \dfrac{1}{2}mv^2.$, $KE = 0.5(30.0 \, kg)(0.500 \, m/s)^2,$, $KE = 3.75 \, kg \cdot m^2/s^2 = 3.75 \, J$. Watch the recordings here on Youtube! The person actually does more work than this, because friction opposes the motion. The calculated total work $$W_{total}$$ as the sum of the work by each force agrees, as expected, with the work $$W_{net}$$ done by the net force. The net work $$W_{net}$$ is the work done by the net force acting on an object. Legal. The net force arises solely from the horizontal applied force $$F_{app}$$ and the horizontal friction force $$f$$. It is moved to a point B. (b) A person holding a briefcase does no work on it, because there is no motion. In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. Find Enthalpy change if ΔU is 418 J. No energy is transferred to or from the briefcase. A mass of 10 kg is at a point A on a table. Calculate the work done in the following reaction when 2 mol of NH4NO3 decomposes at constant pressure at 10o °C. Moreover, they are also equal in magnitude and opposite in direction so they cancel in calculating the net force. Example $$\PageIndex{3}$$: Determining Speed from Work and Energy. Calculate pressure-volume type work and ΔU. As expected, the net work is the net force times distance. Figure 1. examples of work. Furthermore, $$W_{fr} = df' \, cos \, \theta = - Fd'$$, where $$d'$$ is the distance it takes to stop. Example 2: Polytropic work A gas in piston‐cylinder assembly undergoes a polytropic expansion. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. The answers depend on the situation. The net work on a system equals the change in the quantity $$\frac{1}{2}mv^2$$. Such a situation occurs for the package on the roller belt conveyor system shown in Figure. Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). 4 HCl(g)  + O2(g)  →  2 Cl2(g)   +  2 H2O(g), The reaction is 4 HCl(g)  + O2(g)  →  2 Cl2(g)   +  2 H2O(g), Given 2 moles of HCl are used, hence dividing equation by 2 to get 2 HCl, we get, 2 HCl(g)  + ½O2(g)  →   Cl2(g)   +  H2O(g), Δn = nproduct (g)   – nreactant (g) = (1 + 1) – (2 + ½) = 2 – 5/2 = – ½, Work done in chemical reaction is given by, ∴ W = – Δn RT = – (-½) mol × 8.314 J K-1 mol-1 × 423 K = 1758 J, Positive sign indicates that work is done by the surroundings on the system, Ans: Work done by the surroundings on the system in the reaction is 1758 J. When force acts in direction of motion of body, it is called positive work, Direction of force and Direction of Motion in the same direction, When force acts opposite to direction of motion of body it is called negative work, Direction of force and Direction of Motion are at angle of 180 degrees, When force acts perpendicular to direction of body Or when there is no displacement, Sometimes force acts perperndicular to the direction of body, So, angle made between force and distance is 90 degree, Here, Force is appled in vertical direction, But luggage is carried in horizontal direction, Since angle between force and distance is 90 degree, Work Done when Force Acts opposite to Direction of Motion, When force acts opposite to direction of motion, Angle made between direction of force and direction of motion is 180 degree, So, in this case, As can be seen from the picture (p-V diagram), work is path dependent variable. What is ΔU? is done. You can see that the area under the graph is $$F \, cos \, \theta$$, or the work done. a) If volume remains constant, what is ΔU? Does it remain in the system or move on? We are aware that it takes energy to get an object, like a car or the package in Figure, up to speed, but it may be a bit surprising that kinetic energy is proportional to speed squared. c) Suppose that as the original sample absorbs heat, it expands against atmospheric pressure and does 600 kJ of work on its surroundings. 7.2: Kinetic Energy and the Work-Energy Theorem, [ "article:topic", "work-energy theorem", "authorname:openstax", "Kinetic Energy", "net work", "license:ccby", "showtoc:no", "program:openstax" ], Dynamics: Force and Newton's Laws of Motion, Motion Equations for Constant Acceleration in One Dimension, Creative Commons Attribution License (by 4.0). Required fields are marked *, on Problems on Internal Energy Change and Enthalpy Change. Have questions or comments?