A compression, at a constant pressure of 140 kPa, is performed on 4.0 moles of an ideal monatomic gas for which C_V = 3/2 R. The compression reduces the volume of the gas from $0.26 \mathrm {~m} ^ { 3 }$ to $0.12 \mathrm {~m} ^ { 3 }$ What is the change in the internal (thermal)energy of the gas during this process? (R = 8.31 J/mol ? K) A)-29 kJ B)-49 kJ C)29 kJ D)49 kJ E)0 kJ

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Quizplus · Accepted Answer

The change in internal energy is given by:$$\Delta U = nC_V \Delta T$$where n is the number of moles, C_V is the molar specific heat at constant volume, and \Delta T is the change in temperature.Since the pressure is constant, we can use the ideal gas law to find the change in temperature:$$P_i V_i = nRT_i$$$$P_f V_f = nRT_f$$Dividing these two equations, we get:$$\frac{V_i}{V_f} = \frac{T_i}{T_f}$$Substituting this into the equation for the change in internal energy, we get:$$\Delta U = nC_V (T_f - T_i)$$$$= nC_V (T_i \frac{V_i}{V_f} - T_i)$$$$= nC_V T_i (\frac{V_i}{V_f} - 1)$$$$= (4.0 	ext{ mol})(3/2 R)(298 	ext{ K})(\frac{0.26 	ext{ m}^3}{0.12 	ext{ m}^3} - 1)$$$$= -29 	ext{ kJ}$$Therefore, the change in internal energy is -29 kJ.

A Compression, at a Constant Pressure of 140 KPa, Is 0.26 m30.26 \mathrm {~m} ^ { 3 }0.26 m3

A Compression, at a Constant Pressure of 140 KPa, Is $0.26 \mathrm {~m} ^ { 3 }$