KCL is associated with_________

Options

A. Mesh analysis
B. Nodal analysis
C. Both mesh and nodal
D. Neither mesh nor nodal

Explanation

We have `pV=(m)/(M)RT`
`or, (rho)=(m)/(V)=(Mp)/(RT) …(i)`
The dew point is `16.1^(@)C`and the saturation vapour pressure is 13.6mm of mercury at the dew point. This means that the present vapour pressure is `13.6mm` of mercury.
At this pressure and temperature, the density of vapour will be
`(rho)=(Mp)/(RT)`
`((18g mol^(-1))(13.6xx10^(-3)m)(13600kgm^(-3))(9.8m s^(-2)))/((8.3JK^(-1) mol^(-1))(300K))`
`=13.gm^(-3)`.
Thus, 1 litre of moist air at `27^(@)C `contains 0.0131g of vapour.
The pressure of dry air at `27^(@)C` is d`753.6mm- 13.6mm=740mm of mercury. THe density if air at STP is `0.001293g(cc)^(-1)`. The density at `27^(@)C` is given by equation(i),
`((rho)_(1))/((rho)_(2))=(p_(1)/T_(1))/(p_(2)/T_(2))`
or, `((rho)_(@))=(p_(2)T_(1))/T_(2)P_(1))(rho)_(1)`
`=(740xx273)/(300xx760)xx0.001293g(cc)^(-1)`
`=.001457g(cc)^(-1)`
Thus, 1 litre of moist air contains 1.145g of dry air. the mass of 1 litre of moist air is `1.1457g+0.0131g ~~1.159g.`