Please answer both! 5. A sample of blood plasma occupies 0.550 dm^3 at 0 degrees
ID: 1027675 • Letter: P
Question
Please answer both!
5. A sample of blood plasma occupies 0.550 dm^3 at 0 degrees C and 1.03 bar, and is compressed isothermally by 0.57 percent by being subjected to a constant external pressure of 95.2 bar. Calculate the magnitude of the work done.
6. A strip of magnesium metal of mass 12.5 g is dropped into a beaker of dilute hydrochloric acid. Given that the magnesium is the limiting reactant, calculate the work done by the system as a result of the reaction. The atmospheric pressure is 1.00 atm and the temperature 20.2 degrees C.
Explanation / Answer
5) The initial volume of the sample of blood plasma is Vi = 0.550 dm3 at 0°C. The isothermal compression of the sample results in lowering of the volume by 0.57%; therefore, the final volume of the sample is Vf = [(0.550) – 0.57%of(0.550)] dm3 = [(0.550) – (0.57/100)*(0.550)] dm3 = 0.546865 dm3.
The constant external opposing pressure is Pext = 95.2 bar; therefore, the work done is W = -Pext*(Vf – Vi) = -(95.2 bar)*(0.546865 – 0.550) dm3 = 0.2899875 bar.dm3 0.290 bar.dm3 (ans).
6) Write down the balanced chemical equation for the reaction.
Mg (s) + 2 HCl (aq) ---------> MgCl2 (aq) + H2 (g)
As per the stoichiometric equation,
1 mole Mg = 1 mole H2.
Atomic mass of Mg = 24.305 g/mol; therefore, mole(s) of Mg corresponding to 12.5 g Mg = mole(s) of H2 generated = (12.5 g)/(24.305 g/mol) = 0.5143 mole.
Assume the H2 produced to behave like an ideal gas. The temperature of the system is 20.2°C, i.e, T = (20.2 + 273) K = 293.2 K. Use the ideal gas law to find out the volume of the gas produced. Put P = 1.00 atm in the gas law.
P*V = n*R*T
=====> (1.00 atm)*V = (0.5413 mole)*(0.082 L-atm/mol.K)*(293.2 K)
=====> (1.00 atm)*V = 13.0141 L-atm
=====> V = 13.0141 L.
The work done is W = -Pext*V = -(1.00 atm)*(13.0141 L – 0.0 L) = -13.0141 L-atm -13.01 L-atm (ans).