Answer :
Final answer:
The value of w, when 1.35 kg of H2O(l), initially at 25.0 ºC, is converted into water vapor at 173 ºC against a constant external pressure of 1.00 atm, is approximately -26,461 J.
Explanation:
To find the work done (w) when 1.35 kg of water is converted into water vapor against a constant external pressure of 1.00 atm, we can use the formula:
[tex]\[w = -P(V_f - V_i)\][/tex]
Where:
- w is the work done (in Joules, J)
- P is the external pressure (in atmospheres, atm)
- V_f is the final volume (in liters, L)
- V_i is the initial volume (in liters, L)
First, we need to calculate the initial and final volumes of water. We are given that the density of liquid water is 1.00 g/mL, which is equivalent to 1.00 kg/L. Therefore, the initial volume (V_i) can be calculated as:
[tex]\[V_i = \frac{mass}{density} = \frac{1.35 kg}{1.00 kg/L} = 1.35 L\][/tex]
Next, we need to calculate the final volume (V_f). Since the water is being converted into water vapor, we can use the ideal gas law:
[tex]\[PV = nRT\][/tex]
Where:
- P is the pressure (1.00 atm)
- V is the volume (in liters)
- n is the number of moles
- R is the universal gas constant (8.314 J/(mol·K))
- T is the temperature (in Kelvin, K)
We need to convert temperatures to Kelvin by adding 273.15:
Initial temperature (25.0 ºC) = 25.0 + 273.15 = 298.15 K
Final temperature (173 ºC) = 173 + 273.15 = 446.15 K
Now, we can find the number of moles using the mass of water and its molar mass:
Molar mass of water (H2O) = 18.015 g/mol = 0.018015 kg/mol
Number of moles =[tex]\(\frac{mass}{molar mass} = \frac{1.35 kg}{0.018015 kg/mol} \approx 74.87 mol\)[/tex]
Now, we can calculate the final volume (V_f) using the ideal gas law:
[tex]\[V_f = \frac{nRT}{P} = \frac{74.87 mol \times 8.314 J/(mol·K) \times 446.15 K}{1.00 atm} \approx 26182.55 L\][/tex]
Now, we can calculate the work done:
[tex]\[w = -P(V_f - V_i) = -1.00 atm \times (26182.55 L - 1.35 L) \approx -26,461 J\][/tex]
So, the work done (w) is approximately -26,461 J.
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