1.

The temperature of 3 kg of nitrogen is raised form `10^(@)C` to `100^(@)C`, Compute the heat added, the work done, and the change in internal energy if (a) this is done at constant volume and (b) if the heating is at constant pressure. For nitrogen `C_(p) = 1400 J kg^(-1) K^(-1)` and `C_(v) = 740 J kg^(-1) K^(-1)`.

Answer» At constant volume `Delta W = 0`.
Now `Delta U` is always given by `Delta U = mC_(v) Delta T`
`= 3 xx 740 xx (100 - 10) = 199800 J`
According to the first law of thermodynamics
`Delta Q = Delta U + Delta W`
`Delta Q = Delta U = 199800 J`
b. At constant pressure `Delta Q = mC_(P) Delta T = 3 xx 1040 xx (100 - 10) = 280800 J`
`Delta U` is always given by `Delta U = ,mC_(v) Delta T = 199800 J`
According to the first law of thermodynamics `Delta Q = Delta U + Delta W`
or `Delta W = Delta Q - Delta U = 280800 - 199800 = 81000 J`


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