3. A cylindrical tank is capable of storing 20,000 foot-pounds [ft lbf] of energ

Question

3. A cylindrical tank is capable of storing 20,000 foot-pounds [ft lbf] of energy for use in your roommate's 2006 Subaru WRX. The tank contains nitrous oxide, N20, which you may assume is an ideal gas and has a molecular weight of 44 grams per mole [g/mol) The tank fits in a space in the trunk that is no wider than 6 inches fin] and no longer than 10 inches [in), and you may assume the warm Clemson weather is keeping the temperature of the tank inside the trunk at a constant 100 degrees Fahrenheit [°F). What is the weight of the nitrous oxide occupying the tank when full, in units of dynes (dynel? (Hint: you may find it helpful to consider the most-overlooked energy relationship, E-PV, in your solution approach)

Explanation / Answer

Energy stored by the tank=20,000 ft lbf *(1.356 J/ft lbf)=27120 J=27120 J*(1*10^7 erg/J)=27120 *10^7 erg (where erg =1 dyne/cm^2)

Energy of gas=E=Pressure(P)-volume(V) work done by the gas=P*V...............(1)

As the gas is ideal so volume can be calculated using ideal gas equation,

PV=nRT (n=mol of gas=mass of gas/molecular weight=m/44g/mol

R=universal gas constant=0.0821L atm/K.mol

T=temperature=100 deg F=(100-32)*5/9=37.8 deg C=37.8+273=310.8 K

V=nRT/P=(m/44g/mol)*(0.0821L atm/K.mol)*(310.8K)/P=(0.580 *m/P)(g L atm)

or, V=(0.580 *m/P)(g L atm).....................(2)

Plug eqn(2) in (1)

E=PV =(P)((0.580 *m/P)(g L atm)=((0.580 *m)(g L atm)

or, 27120 J=(0.580 *m)(g L atm)

or, 27120 J=(0.580 *m)(g L atm)

[R=8.314 J/K.mol=0.0821L atm/K.mol, 1Latm=8.314 J/0.0821=101.267 J]

So,27120 J=(0.580 *m)(g L atm) *(101.267 J/Latm)=m*58.735 gJ

or,27120 J=m*58.735 gJ

m=461.737 grams

mass or force(in dyne)=mass *g ,(g=acceleration due to graviy=980cm/sec^2)

mass (in dyne)=461.737 g*(980cm/sec^2)=452502.418 dynes

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