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A gas expands from 3 dm^(3) to 5 dm^(3) ...

A gas expands from `3 dm^(3)` to `5 dm^(3)` against a constant pressure of 3 atm. The work done during expansion is used to heat 10 mole of water at temperature 290 K. Calculate final temperature of water. Specific heat of water = `4.184 J//g//K`.
(given : 1 atm = 101.33 J)

Text Solution

Verified by Experts

The correct Answer is:
290.81 K
Work done against external pressure and thus proces is irreversible
`w=- PDeltaV`
`DeltaV = (5-3) = 2dm^(3) = 2 xx 10^(-3) m^(3)`
`P=3atm = 3 xx 1.013 xx 10^(5) Nm^(-2)`
`w=-3 xx 1.013 xx 10^(5) xx 2 xx 10^(-3) = -607.8` J
Since, the work is used in heating water and thus
`-w = q= n xx C xx DeltaT`
Or `607.8 = 10 xx 4.184 xx 18 xx DeltaT`
`DeltaT = 0.81`
Final temperature =`290 + 0.81 = 290.81` K
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