Question

3. Nitrogen gas is stored at a temperature of 200 K and a specific volume of 0,004 m3/kg. Determine the pressure on the basis of (a) the ideal gas equation of state (b) assuming a real gas behavior with the compressibility factor 0.8 (c) the van der Waals equation of state (Hint: Critical point properties of Nitrogen from Table A-1 (page 882) are: R=0.2968 kJ/kg, To - 126.2 k, P. - 3.39 Mpa. Using these values, a. 27R'T/64P,-0.175 m.kPa/kg?bRT/8P - 0.00138 m/ke) 4. A block of 10 kg iron at 300°C iron and a block of 20 kg copper at 200°C are dropped in a bucket of water holding 200 kg of water at 40C. 4000 kb of heat is lost from the water to the environment. What is the final temperature of the iron and copper block and the water after thermal equilibrium is reached. The specific heat of iron, copper, and water are 0.45 kike K 0.403 kJ/kg. K and 4.22 kJ/kg.K respectively 5. The manometer shown in the figure was used in two different cities. The specific gravity of the fluid in the manometer in the second city was three times the specific gravity of the fluid used in the first city. The atmospheric pressure in the second city is also three times the atmospheric pressure of the first city. (a) The pressure of the tank in the first city is: (i) Three times (ii) same (ili) one third of the pressure in the second city (b) The manometer fluid height, h in the second city is (0) Three times (i) same one third of the height of the in the second city

Answer 1

given T = 200k Bpeuifie volume , V = 0.000 m @ Ideal gas equation PV = RT - volume - 12 given per kg we should of use specific universal gas gas constant in place constant. specific gas - MWAMolecular weight. Nitrogen molewlar weight = 28 kg kmol. Os) i. Pa 8.314 (kPara I x 200 20 0,004 * p 59.3852 kPa 0.004 P = 14846.42 kPa | 6 real gas behariour with 2:0.8 PV = zB el hot

0,8{ $3.]* 200 0.004 47.5085 0.004 IP = 11877.142 kPa o vanderwall's equation of state R = 0.2968 KSK Kak Tor = 126.8k Por = 3,39 MPa = 3390 kPa a = 21. RTB a= a 1 0.29682 68 x 126.92 3390 a = 0.1459 mai mare 0.2968 x 126.2 • a * 3,390 b = 0.00138 mblkg,

vanderwalls equation of state (P+ ) (vo-b) - RT (P + 2) = Bob pe (01) - 1.0 P= 60:2008 Ave) - ( ) 0.2968 goo 0.175 0.004 -0.00138) po 226 56.488 – 10937.5 P = 1118.988 kPa
yI have shown the usage of specific gas constant in the first two parts, you can directly use the value of R given in the 3rd Subpart also.both will give same answer.I have also recalculated the vanderwalls constants to cross check the values.Thank you.