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The volume of steam produced by 1g of water at 100°c is 1650 cm3.Calculate the change in internal energy during the change of state givenJ = 4.2 × 107 erg cal–1 g = 98 J cm/s2latent heat of steam = 540 cal/g

Vaibhav shukla , 7 Years ago
Grade 11
anser 3 Answers
Arun

Last Activity: 7 Years ago

Dear Vaibhav
 
Mass of water = 1 gm = .001 kg
Volume of water = mass/density = .001/1000=.000001 m³
= 1 cm³
Change in volume = 1650 -1 = 1649 cm³
dQ = m*L = 540 * 4.2 * 10^7 erg.
P = 1 atm = 76* 13.6 *9.8
dU = dQ - P* dV
 = 22.68 * 10^9 - 1.67 * 10^9
 = 21.01 * 10^9  erg
 
Regards
Arun(askIITians forum expert)
 

ankit singh

Last Activity: 4 Years ago

From first law of thermodynamics, 

δQ=dU+δW
54×4.184=dU+PdV
54×4.184=dU+1.013×105×167.1×10−60
On solving, we will get dU=208.7J

Ravindra Nagalapura

Last Activity: 2 Years ago

Not correct , initial value of steam = 0, since there is no state change happened as yet.
so change in volume = (1650 – 0) = 1650 cm3

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