Task 1) An energy of 2145J was added from the system, calculate the internal energy if the system performs 1165J of work. 2) Determine the change in internal energy of the system if 97.5 J of heat energy were removed from it and if 42.25 J work was done on the system. 3) A 1987.3J of energy was removed from a gas. The gas was compressed at a pressure of 890Pa from initial volume of 6 m^3 to final volume of 3.5 m^3. Calculate the WORK done and the change in Internal Energy. 4) 3520.9J of heat was added to the system. The gas expand from 0.5 m^3 to 1.23 m^3 at a pressure of 43000Pa. Calculate the WORK done and CHANGE IN INTERNAL ENERGY.
Answer (1)
We'll use the first law of thermodynamics: ΔU = Q - W, where: - ΔU = change in internal energy - Q = heat added to the system (positive) or removed from the system (negative) - W = work done by the system (positive) or on the system (negative) 1) An energy of 2145 J was added from the system, calculate the internal energy if the system performs 1165 J of work. - Q = +2145 J (heat added) - W = +1165 J (work done by the system) ΔU = Q - W = 2145 J - 1165 J = 980 J The change in internal energy is 980 J. 2) Determine the change in internal energy of the system if 97.5 J of heat energy were removed from it and if 42.25 J work was done on the system. - Q = -97.5 J (heat removed) - W = -42.25 J (work done on the system) ΔU = Q - W = -97.5 J - (-42.25 J) = -55.25 J The change in internal energy is -55.25 J. 3) A 1987.3 J of energy was removed from a gas. The gas was compressed at a pressure of 890 Pa from an initial volume of 6 m³ to a final volume of 3.5 m³. Calculate the WORK done and the change in Internal Energy. - Q = -1987.3 J (heat removed) - W = PΔV = 890 Pa * (3.5 m³ - 6 m³) = 890 Pa * (-2.5 m³) = -2225 J (work done on the gas during compression) ΔU = Q - W = -1987.3 J - (-2225 J) = 237.7 J Work done = -2225 JChange in Internal Energy = 237.7 J 4) 3520.9 J of heat was added to the system. The gas expands from 0.5 m³ to 1.23 m³ at a pressure of 43000 Pa. Calculate the WORK done and CHANGE IN INTERNAL ENERGY. - Q = +3520.9 J (heat added) - W = PΔV = 43000 Pa * (1.23 m³ - 0.5 m³) = 43000 Pa * 0.73 m³ = 31390 J (work done by the gas during expansion) ΔU = Q - W = 3520.9 J - 31390 J = -27869.1 J Work done = 31390 JChange in Internal Energy = -27869.1 J Remember that work is positive when done by the system (expansion) and negative when done on the system (compression). Heat is positive when added and negative when removed.
