Calculate the amount of pv work done on the surroundings when 1.00 kg of h2o absorbs energy from sunlight and vaporizes at 1.0 atm and 25 degrees c. assume the volume of liquid h2o is negligible (~0 m^3) compared to that of vapor. (r = 8.314 j / k-mol)

137639.472 J

Explanation:

Given, Mass of water = 1 kg = 1000 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

Thus,

Pressure = 1.0 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

Using ideal gas equation as:

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1.0 atm × V = 55.508 mol × 0.0821 L.atm/K.mol × 298.15 K  

⇒V = 1358.7312 L

The expression for the calculation of work done by the surroundings is shown below as:

Where, P is the pressure

is the change in volume

From the question,  

= 1358.7312 - 0 L = 1358.7312 L

P = 1.0 atm

Also, 1 atmL = 101.3 J

So,  

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

50 inches should be the resultant vector