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The volume of steam produced by 1g of wa...

The volume of steam produced by `1g` of water at `100^(@)C` is `1650 cm^(3)`. Calculate the change in internal energy during the change of state. Given `J= 4.2xx10^(7) erg`.
`cal.^(-1), d= 981 cm^(s-2)`. Latent heat of stream `= 540 cal. G^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
`2.1`
Here `J = 4.2xx10^7` erg `"cal"^(-1)`
Latent heat of steam, `L = 540 "cal g"^(-1)`
Mass of water = 1 g
Temperature of water = `100^@C`
Initial volume, `V_1 = 1cm^3`
Final volume, `V_2 = 1650 cm^3`
`:.` Change in volume, `dV = V_2 - V_1 = 1650 - 1=1649 cm^3`
When 1g of water at `100^@C` is changed to steam at `100^@C` temperature remains constant. So the heat supplied is dQ = mL `= 1xx 540 = 540 cal = 540 xx 4.2xx10^7` erg
Pressure, P = 1 atm = `76 xx 13.6 xx 981 "dyne cm"^(-2)`.
From first law of thermodynamics, dU = dQ - PdV
`= 540 xx 4.2 xx10^7 - 76xx13.6 xx 981xx1649`
` = 22.68 xx 10^9 -1.67xx10^9`
`= 21.01xx10^9 = 2.1xx10^(10)` erg
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