In this problem, a monatomic ideal gas undergoes three reversible processes: doubling of volume at constant pressure, doubling of pressure at constant volume, and returning to the initial state along a curved path. The work effects for processes a and b are zero, while the work effect for process c is given by W = -6.643 × 10⁻⁴V³ + 0.6667V. The heat effects for processes a and b are also zero, and the heat effect for process c is equal to the work effect.
In this problem, we have a monatomic ideal gas undergoing three reversible processes.
Process a:
In process a, the volume of the gas doubles at constant pressure. Since the pressure is constant, the work done by the gas is zero. To calculate the heat effect, we can use the equation Q = ΔU + W, where Q is the heat, ΔU is the change in internal energy, and W is the work done on or by the gas. Since the internal energy depends only on the temperature for an ideal gas, the change in internal energy is given by ΔU = nCvΔT, where n is the number of moles of gas, Cv is the molar heat capacity at constant volume, and ΔT is the change in temperature. In this case, ΔU = 0 since the temperature is constant, so the heat effect Q = 0.
Process b:
In process b, the pressure of the gas doubles at constant volume. Again, the work done by the gas is zero since the volume is constant. The heat effect can be calculated using the same equation Q = ΔU + W. Since the internal energy depends only on the temperature, the change in internal energy is zero, so the heat effect Q = 0.
Process c:
In process c, the gas returns to its initial state along a curved path described by the equation P = 6.643 × 10⁻⁴V² + 0.6667. To calculate the work effect, we can use the equation W = ∫PdV, where P is the pressure and V is the volume. Evaluating the integral, we find that the work done on the gas is given by W = -6.643 × 10⁻⁴V³ + 0.6667V. To calculate the heat effect, we use Q = ΔU + W. Since the process is reversible, the change in internal energy ΔU = 0, so the heat effect Q = W.