2.21已知水在100℃,101.325kPa下的摩尔蒸发烙 Delta (a)_(mp)(H)_(m)=40.668kJcdot mo(l)^-1 试分别计算下列两-|||-过程的Q,W, △U 及 Delta (H)_(0) (水蒸气可按理想气体处理)-|||-(1)在100℃,101.325 kPa条件下,1kg水蒸发为水蒸气;-|||-(2)在恒定100℃的真空容器中,1 kg水全部蒸发为水蒸气,并且水蒸气压力恰好为-|||-101.325kPa。

步骤 1：计算1kg水的摩尔数
1kg水的摩尔数为：$n = \frac{1000}{18.02} = 55.51 \text{ mol}$。
步骤 2：计算过程(1)的Q, W, $\Delta V$ 及 $\Delta H$
- Q = $\Delta H = n \times \Delta_{mp}H_m = 55.51 \times 40.668 \times 10^3 = 2.2568 \times 10^6 \text{ J}$
- W = $-P_{amb} \times \Delta V = -101.325 \times 10^3 \times (V_{vapor} - V_{water}) \approx -101.325 \times 10^3 \times V_{vapor} = -101.325 \times 10^3 \times \frac{n \times R \times T}{P} = -101.325 \times 10^3 \times \frac{55.51 \times 8.314 \times 373.15}{101.325 \times 10^3} = -1.7216 \times 10^5 \text{ J}$
- $\Delta V = V_{vapor} - V_{water} = \frac{n \times R \times T}{P} = \frac{55.51 \times 8.314 \times 373.15}{101.325 \times 10^3} = 1.700 \times 10^{-2} \text{ m}^3$
- $\Delta H = 2.2568 \times 10^6 \text{ J}$
步骤 3：计算过程(2)的Q, W, $\Delta V$ 及 $\Delta H$
- W = $-P_{amb} \times \Delta V = 0$ (因为$P_{amb} = 0$)
- $\Delta H = 2.2568 \times 10^6 \text{ J}$ (因为始末态与(1)相同)
- $\Delta U = 2.08464 \times 10^6 \text{ J}$ (因为始末态与(1)相同)
- Q = $\Delta U - W = 2.08464 \times 10^6 - 0 = 2.08464 \times 10^6 \text{ J}$