Question

Определите во сколько раз увеличился объем системы при полном испарении 1 л жидкой воды при постоянной температуре 100°с и постоянном давлении 1 атм, помогите, пожалуйста!!!!

Answer 1

Дано
H2O
V0=1 L=1000 м3
t=100 c, T=t+273 = 373K
P=1атм = 101.325kПа
найти V-?
находим массу воды :р(Н2О)=1 г/см3
m(H2O)=V(H2O)*р(H2O)=1000*1=1000 g
Решаем по уравнению Менделеева - Клайперона
P*V=(m/M)*R*T, R=8.31 Дж , M(H2O)=18 g/mol
V=(m*R*T) / (P*M)=(1000*8.31*373) / (101.325*18) = 1700 L
V/V0=1700 / 1 = 1700 раз
ответ 1700 раз

Answer 2

Calculating the Increase in Volume upon Complete Evaporation of 1 L of Water

To determine the increase in volume upon complete evaporation of 1 L of liquid water at a constant temperature of 100°C and a constant pressure of 1 atm, we can use the ideal gas law and the concept of molar volume.

The molar volume of an ideal gas at standard temperature and pressure (STP) is approximately 22.4 liters per mole. This means that one mole of an ideal gas occupies 22.4 liters at a temperature of 0°C and a pressure of 1 atmosphere.

Using this information, we can calculate the increase in volume when 1 mole of water vaporizes at 100°C and 1 atm pressure.

Step 1: Convert 1 liter of water to moles The molar mass of water (H2O) is approximately 18.015 grams per mole. Therefore, 1 liter of water is approximately 55.56 moles (1000 grams / 18.015 grams/mole).

Step 2: Calculate the increase in volume When 1 mole of water vaporizes, it occupies approximately 22.4 liters of space at STP. Therefore, 55.56 moles of water vapor would occupy: 55.56 moles * 22.4 liters/mole = 1244.704 liters

So, the volume of the system would increase by approximately 1244.704 times when 1 liter of liquid water completely evaporates at a constant temperature of 100°C and a constant pressure of 1 atm.