Calculate the work done against external atmospheric pressure when 1 g of water changes to
1672cm3
of steam. Take the atmospheric pressure as
1.013×105Nm−2
P deltaV= ....
1.522j
To calculate the work done against external atmospheric pressure when water changes to steam, we can use the formula:
Work = Pressure * Change in Volume
Given:
Mass of water (m) = 1 g
Change in volume (ΔV) = 1672 cm^3
Atmospheric pressure (P) = 1.013 × 10^5 N/m^2
First, we need to convert the mass of water to the volume of water. We can do this using the density of water.
The density of water is approximately 1 g/cm^3.
Volume of water (V) = Mass / Density = 1 g / 1 g/cm^3 = 1 cm^3
Now, we can calculate the work done against atmospheric pressure:
Work = Pressure * Change in Volume
= (1.013 × 10^5 N/m^2) * (1672 cm^3 - 1 cm^3)
To get the answer, we need to convert the units of volume from cm^3 to m^3 and N to J.
Note that 1 m^3 = 10^6 cm^3 and 1 N = 1 J.
Work = (1.013 × 10^5 N/m^2) * ((1672 cm^3 - 1 cm^3) / 10^6)
Now, let's calculate:
Work = (1.013 × 10^5 N/m^2) * (1672 / 10^6 m^3)
= 1.013 × 10^5 N/m^2 * 1.672 × 10^-3 m^3
= 1.695936 × 10^2 J
Therefore, the work done against external atmospheric pressure when 1 g of water changes to 1672 cm^3 of steam is approximately 169.59 J.