Beneath the moveable piston in the drawing,2.25 moles of a monatomic ideal gas is enclosed at 314 K.The initial volume of the gas is 3.0 m³.The gas is compressed isothermally to a final volume of 1.0 m³.How much heat is removed from the gas during this process?

Question

Quizplus · Accepted Answer

Since the process is isothermal, the temperature remains constant at 314 K. Therefore, the initial pressure of the gas can be calculated using the ideal gas law:
PV = nRT
P = nRT/V = (2.25 mol)(8.31 J/mol*K)(314K)/(3.0m³) = 605.01 Pa
The work done on the gas during the compression can be calculated using the formula for the work done in an isothermal process:
W = nRT ln(Vf/Vi) = (2.25 mol)(8.31 J/mol*K)(314K) ln(1.0m³/3.0m³) = -6450 J
The negative sign indicates that work is done on the gas (i.e., the gas is compressed).
Since the process is isothermal and no other forms of energy are involved, the heat removed from the gas is equal to the work done on the gas:
ΔQ = W = -6450 J
Therefore, the correct answer is (A): -6450 J.

Beneath the Moveable Piston in the Drawing,2