Question: Problem 2 : A long rigid cylindrical container ( a tube ) of diameter D = 5 c m and length, L = 1 0

Problem 2: A long rigid cylindrical container (a tube) of diameter D=5cm and length, L=100cm contains 0.1 kg of gas the pressure, p of which is related to its temperature, T and specific volume, v by the following function:
p(v,T)=RTv-b-av(v+b)T2(I)
where R,a and b are constants.
a) What are the base dimensions and base SI units of the constants R,a and b?[6]
b) Show that Jkg=m2s2.
c) Calculate the value of a and b in base SI units if
a=0.4278R2Tc25pc and b=0.0867RTcpc
where pc=38atm and Tc=135K are the so-called critical pressure and temperature,
respectively and the value of R in base SI units is 189.
d) Convert the constants R,a, and b to US customary units.
e) Using (I) calculate the gas temperature if the pressure is 21 atm .
f) More of the same gas is added to the rigid container until the pressure and temperature
are 25 atm and 250 K , respectively. Using equation, (I) determine the exact amount of gas mass added in grams. (Exact means that we CANNOT use an iterative procedure or use graphical methods/root-finder programs via our calculator because both are approximations. To find the exact amount we have to find exact expressions for the roots of the polynomial represented by equation (I).)
g) For the situation depicted by part f), evaluate dpdT|T=300x in SI units.
h) Develop an expression for v12pdv in terms of u1,v2 and T for an isothermal process. Then evaluate the integral's value if v1=1.610-3m3kg,v2=210-3m3kg and T=300K.
Problem 2 : A long rigid cylindrical container (

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