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Nitrogen at 100 kPa and 25°C in a rigid container is heated (isochoric process) until its pressure is 270 kPa. Calculate the work done and the heat transferred during this process, in kJ/kg. For the nitrogen, cv = 0.743 kJ/kg·K at room temperature. (Round the final answers to one decimal place.)

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MechEngineer

Answer:

376.4 kJ/kg

Explanation:

T = 25 C = 25 + 273 = 298 K

Since this is an isochoric process, the volume stays constant. The heat transfer comes mainly from internal energy, or change in term of change in temperature.

Assume idea gas law, we have

[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex]

[tex]\frac{100}{298} = \frac{270}{T_2}[/tex]

[tex]T_2 = 270\frac{298}{100} = 805.6 K[/tex]

So the heat transfer due to change of temperature is

[tex]\dot{E} = c_v\Delta T = c_v (T_2 - T_1) = 0.743 * (805.6 - 298) = 376.4 kJ/kg[/tex]

akande212

Answer:

In an isochoric process the workdone is zero this is because there is no volume change and when there is no volume change no work is done.

The heat transfered is 376.7kJ/kg.

Explanation:

From first law of thermodynamics

Q = U + W

Since W = 0 then Q = U = Cv(T2 - T1)

T2 is unknown T1 = (25 +273)K =298K

P1 = 100kPa P2 = 270kPa

P1 / T1 = P2 /T2

T2 = P2 ×T1 / P1 = 270 × 298 / 100 = 805K

THEREFORE,

Q = 0.743×(805 - 298)

= 376.7kJ/Kg of heat.

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