Question

Three moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 91.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 182 J. Three paths are plotted on a PV diagram, which has a horizontal axis labeled V (liters), and a vertical axis labeled P (atm). The green path starts at point I (0.300,2.00), extends vertically down to point A (0.300,1.50), then extends horizontally to point F (0.800,1.50). The blue path starts at point I (0.300,2.00), and extends down and to the right to end at point F (0.800,1.50). The orange path starts at point I (0.300,2.00), extends horizontally to the right to point B (0.800,2.00), then extends vertically down to end at point F (0.800,1.50). (a) For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas. WIAF = J WIBF = J WIF = J (b) For the paths IAF, IBF, and IF in the figure above, calculate the net energy transferred to the gas by heat in the process. QIAF = J QIBF = J QIF = J

Answer 1

The work done on the gas and the heat transferred in the thermodynamic processes can be found by integrating the PV curve or applying the first law of thermodynamics, respectively. Without a specific functional relationship for path IF, the work done cannot be calculated for that path.

Step-by-step explanation:

Work Done and Heat Transfer during Thermodynamic Processes

To calculate the work done on the gas for paths IAF, IBF, and IF, we must consider the area under the curve in the PV diagram for each path. Work done on the gas in a PV diagram is the area under the curve where volume changes at constant pressure, given by the formula W = -PΔV (work is considered positive when done on the system).

Path IBF: The work on the gas during IB is WIB = -P(I)ΔV(IB) = -2.00 atm × (0.800 L - 0.300 L), and no work is done during BF since it's isochoric.

Path IF: It's a direct path and would typically involve an integral to calculate the exact work, but since we lack a functional relationship between P and V, we cannot calculate it with the given data.

The heat exchanged, Q, can be found using the first law of thermodynamics, ΔU = Q - W, where ΔU is the change in internal energy and W is the work done on the gas. Since we know ΔU for the transition from initial to final state, and we've calculated W for paths IAF and IBF, we can determine QIAF and QIBF. For path IF, without a functional relationship between P and V, we cannot determine the work and thus cannot accurately calculate the heat transferred, QIF.

Note: To resolve these equations, we must convert the pressure from atm to J/L by using the conversion factor 1 atm = 101.325 J/L, and then calculate the work and heat values in joules.