Question

Calculate q, w, and Δ when 1.00 mol of water is heated from 0°C to 100°C at a fixed pressure of 1 atm.

Densities of water are 0.9998 g/cm3

at 0°C and 0.9854 g/cm3

at 100°C​

Answer 1

Answer:

sorry

Explanation:

,

,

,and

headache

Answer 2

Answer:

q = 1800 cal

w = -4,36 x10^-3 cal

ΔU = 1799,0 ≈ 1800 cal

Explanation:

I had the same problem studying in Levine's physical chemistry book, the problem is that the answers do not match because they approximate the values ​​too much or they are very small numbers I think:

So de easier way to start is with q

q = mcΔT

q = ( 18g ) ( 1 cal /g °C ) ( 100 °C)

q = 1800 cal

w = - ∫ PdV

w = - 0,18 atm cm³ × $(\frac{1,987 cal/mol K}{82,06 cm^{3}/mol K })$ = - 4,36 x10^-3 cal

That fraction would be like saying:

$\frac{R}{R} = 1$  ( gas constants)

It is to be able to convert the result to calories, if you pay attention to the units you will see that all are canceled except for the calories.

then:

ΔU = q + w

ΔU = 1799,0 ≈ 1800 cal