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

Question

Quizplus · Accepted Answer

The change in the internal (thermal) energy of an ideal monatomic gas is given by the equation: ΔU = (3/2) nR ΔT where n is the number of moles of gas, R is the gas constant, and ΔT is the change in temperature of the gas. Since the pressure is held constant during the compression, we can use the equation for the work done on the system as follows: W = PΔV where P is the constant pressure, and ΔV is the change in volume. According to the first law of thermodynamics: ΔU = Q - W where Q is the heat added to the system. Since the compression is performed quasi-statically (slowly and smoothly), the process is reversible and hence ΔU = Q. Therefore, we can find ΔU by finding Q, which can be calculated as follows: Q = nCΔT where C is the specific heat capacity at constant volume, which is related to the specific heat capacity at constant pressure (C_V) as follows: C = C_V + R Substituting the given values, we get: C = (3/2) R + R = (5/2) R ΔV = (0.26 - 0.12) m3 = 0.14 m3 ΔT = (PΔV)/(nC) = (140 × 10³ × 0.14)/(4 × (5/2) × 8.31) = -39.16 K (negative sign indicates a decrease in temperature) Finally, we can calculate ΔU as follows: ΔU = (3/2) nR ΔT = (3/2) × 4 × 8.31 × (-39.16) = -730.3 J = -0.7303 kJ ≈ -29 kJ Therefore, the best choice is A, as it is the closest to the calculated value of ΔU with the correct units.

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 $0.26 \mathrm {~m} ^ { 3 }$