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Question

A gas expands from a volume of 3.0 dm3 to 5 dm3 against a constant pressure of 3.0 atm. The work done during expansion is used to heat 10.0 mole of water of temperature 290.0 K. Calculate the final temperature of water. (Specific heat of water = 4.184 JK1g1

A
289.193 K
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B
290.807 K
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C
304.526 K
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D
275.470 K
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Solution

The correct option is B 290.807 K
Work done = - P×dV
=-3.0×(5.03.0)
= -6.0 Latm
= -6.0×101.3 J
= -607.8 J
Magnitude of work done = 607.8 J
Consider ΔT be the change in temperature during the process
Heat absorbed = m×s×ΔT
=10.0×18×4.184×ΔT J
According to the given question
Work done = Heat absorbed
607.8 J = m×s×ΔT
= ΔT=607.8m×s
= ΔT=607.810.0×18×4.184
ΔT=0.807K
Hence final temperature = 290 + 0.807
= 290.807 K

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