Question

One gram of water (1 cm3

) becomes 1671 cm3

of steam at a pressure of 1 atm. The

latent heat of vaporization at this pressure is 2256 J/g. Calculate the external work and

the increase in internal energy.​

Answer 1

Answer:

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Answer 2

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GIVEN

• Mass of water (m) = 1 gm

• Initial volume (V1) = 1 cm3

• Final volume (V2) = 1671 cm3

• Pressure(p) = 1 atm

• latent heat (L) = 2256 j/g

FORMULA

• ∆Q= mL

• W = P∆V

• ∆U = ∆Q - W

where,

• ∆Q = Heat added

• W = work done

• ∆U = Internal energy change

• m = mass

• L = latent heat

• p = Pressure

• ∆V = change in Volume

SOLUTION

• ∆V = V2 - V1 = 1671 - 1 = 1670 cm3

so

$\rarr \: w = P∆V = 1 \times 1670 = 1670 \: j$

$\rarr \: ∆Q= mL = 1 \times 1256 = 1256 \: j$

$∆U = ∆Q - W \: = 1256 - 1670 = - 414j$