Answer:
T=655K, W=816kJ/kg
Step-by-step explanation:
consulting the properties of gases, for Helium we have,
![R=2.0769kJ/kg.K,\\c_p=5.1926kJ/kg.K,\\c_v=3.1156kJ/kg.K,](https://img.qammunity.org/2020/formulas/engineering/college/mfwxu39e08xpmet82c92jmqhityfiuvnrw.png)
To find the specific volume we use the equation,
![v=(RT)/(P)](https://img.qammunity.org/2020/formulas/engineering/college/89lmueamb30e6ygnefdjag9uha6rsd8isp.png)
Substituting,
![v=(2.0769*(120+273))/(200*10^3)\\v=4.08m^3/kg](https://img.qammunity.org/2020/formulas/engineering/college/rklyuw8qwlwqbic5y5957vm70wo5zyzqzb.png)
Flow work is gived by,
![W_(flow)=Pv\\W_(flow)=200*10^3*4.08=816kJ/kg](https://img.qammunity.org/2020/formulas/engineering/college/tzizg8tmi1y3qkk0eey28fj1c69a5foljy.png)
To find the temperature of the gas in the tank we need
![u_(tank)=h_(line)\\h_(line)=c_pT_(line)\\u_(tank)=c_vT_(tank)](https://img.qammunity.org/2020/formulas/engineering/college/bsc9mxmdsu5zbvhsoggq1irp82in63fs13.png)
So,
![T_(tank)=(c_p T_(line))/(c_v)\\T_(tank)=(5.1926)/(3.1156)*343\\T_(tank)=655K](https://img.qammunity.org/2020/formulas/engineering/college/js8f20dt1g81q9224wyg4584f3iqyzshb7.png)