A sample of 0.1 g of water at `100^@ C` and normal pressure `(1.013 xx 10^5 Nm^(-2))` requires 54 cal of heat energy to convert to steam at `100^@ C`. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is ____

Options

A. 84.5 J
B. 104.3 J
C. 42.2 J
D. 208.7 J

Answer

D

Explanation

`DeltaQ`=54 cal = `54xx4.18`J =225.72 J …(1)
`DeltaW=PDeltaV`
`=P(V_2-V_1)` `V_2`=volume of steam
`V_1`=volume of water
`=1.013xx10^5[167.1xx10^(-6)- 0.1xx10^(-6)]`
[`because` 1CC=`10^(-6) m^3`]
`=1.013xx10^5xx167xx10^(-6)`
`DeltaW` = 16.917 ..(2)
Now `DeltaU=DeltaQ-DeltaW`
=225.72 - 16.917 [`because` From equ. (1) and (2)]
=208.8 J
`therefore DeltaU` = 208.7 J nearest value