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nevsk [136]
3 years ago
7

1. Nitrogen at an initial state of 300 K, 150 kPa, and 0.2 m3 is compressed slowly in an isothermal process to a final pressure

of 800 kPa. Determine the work done during this process.
Engineering
1 answer:
Sauron [17]3 years ago
7 0

Answer:

Workdone during the process = 130kJ

Explanation:

Workdone by an expanding gas which is also called pressure-volume work, is defined as the pressure exerted by the gas molecules on the walls of the containing vessel. If work done by an expanding gas is the energy transferred to its surroundings i.e If volume increases, workdone is negative(loses energy).

In an isothermal process(constant Temperature of 300K, we use boyles law:

P1V1 = P2V2

V2 = (150 * 0.2)/800

= 0.0375 m3

W = -P(V2 - V1)

= 800*(0.0375 - 0.200)

= 130kJ

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Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee
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Answer:

a) 1 m^3/Kg  

b) 504 kJ

c) 514 kJ

Explanation:

<u>Given  </u>

-The mass of C_o2 = 1 kg  

-The volume of the tank V_tank = 1 m^3  

-The added energy E = 14 W  

-The time of adding energy t = 10 s  

-The increase in specific internal energy Δu = +10 kJ/kg  

-The change in kinetic energy ΔKE = 0 and The change in potential energy  

ΔPE =0  

<u>Required  </u>

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.  

(c)The energy transferred by the heat transfer W in kJ and the direction of  

the heat transfer.  

Assumption  

-Quasi-equilibrium process.  

<u>Solution</u>  

(a) The volume and the mass doesn't change then, the specific volume is constant.

 v= V_tank/m ---> 1/1= 1 m^3/Kg  

(b) The added work is defined by.  

W =E * t --->  14 x 10 x 3600 x 10^-3 = 504 kJ  

(c) From the first law of thermodynamics.  

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

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