An expansion process on an ideal diatomic gas has a linear path between the initial and final states on a pV diagram. The initial pressure is 300 kPa, the initial volume is 0.020 m³, and the initial temperature is 390 K. The final pressure is 160 kPa and the final temperature is 310K. The change in the internal (thermal) energy of the gas is closest to A) -3100 J. B) -1800 J. C) 3100 J. D) 1800 J. E) 0.00 J.

Question

Quizplus · Accepted Answer

The change in internal energy is given by:$$\Delta U = Q - W$$where Q is the heat added to the gas and W is the work done by the gas.For an ideal gas, the heat added is given by:$$Q = nC_v\Delta T$$where n is the number of moles of gas, Cv is the molar specific heat at constant volume, and ΔT is the change in temperature.The work done by the gas is given by:$$W = -P\Delta V$$where P is the pressure and ΔV is the change in volume.Substituting these expressions into the equation for the change in internal energy, we get:$$\Delta U = nC_v\Delta T + P\Delta V$$For a diatomic gas, Cv = 5/2R, where R is the ideal gas constant.Substituting this into the equation for the change in internal energy, we get:$$\Delta U = \frac{5}{2}nR\Delta T + P\Delta V$$Substituting the given values into this equation, we get:$$\Delta U = \frac{5}{2}(1 	ext{ mol})(8.31 	ext{ J/mol K})(310 	ext{ K} - 390 	ext{ K}) + (160 	ext{ kPa})(0.020 	ext{ m}^3 - 0.010 	ext{ m}^3)$$$$\Delta U = -3100 	ext{ J}$$Therefore, the change in the internal energy of the gas is closest to -3100 J.

An Expansion Process on an Ideal Diatomic Gas Has a Linear