Physical Chemistry Chapter 2 Questions/Answers

2.) Calculate the work done by 2.0 mol of a gas when it expands reversibly and isothermally from 1.0 dm³ at 300 K to 3.0 dm³

W=-nRTln(V_f/V_i)

   =-(2.0mol)(8.31)(300)ln(3.0dm³/1.0dm³)

   =-5.5kJ

3.) A sample of methane of mass 4.50 g occupies 12.7 dm³ at 310 K. (a) Calculate the work done when the gas expands isothermally against a constant external pressure of 30.0 kPa until its volume has increased by 3.3 dm³. (b) Calculate the work that would be done if the same expansion occurred isothermally and reversibly.

a.)W=-P_ext ΔV

       =-(30000 Pa)*(3.3dm³)*(1m³/1000dm³)

       =-99 J

b.)W=-nRTln(V_f/V_i)

        =-(4.5g*(16.04 g/mol))*(8.31)*(310K)*ln(16/12.7))

        =-167 J

 4.) In the isothermal reversible compression of 52.0 mmol of a perfect gas at 260 K, the volume of the gas is reduced from 300 cm3 to 100 cm3. Calculate w for this process.

W=-nRTln(V_f/V_i)

   =-(.052mol)*(8.31)*(260K)*ln(100cm³/300cm³)

   =123.4 J

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 is 20.2°C.

Mg_(s) + HCl _(l) --> MgCl_{2 (aq)} + H_{2 (g)}

Moles of H₂ = (12.5 g magnesium)*(1mol/24.305g)*(1mol H₂/1mol Mg)=0.514 mol

Volume of H₂= nRT/P= (0.514 mol)*(8.31)*(293.2 K)/(1 atm)=12.38 L

1 atm=101325 Pa; 1m³=1000L;

W=P_ext* ΔV= (101325 Pa)*(.01238 m³)

=-1.25 kJ

9.) What is the heat capacity of a sample of liquid that rose in temperature by 5.23°C when supplied with 124 J of energy as heat?

C=q/ ΔT=124 J/ 5.23 K

=23.7 J/K

10.) A cube of iron was heated to 70°C and transferred to a beaker containing 100 g of water at 20°C. The final temperature of the water and the iron was 23°C. What is (a) the heat capacity, (b) the specific heat capacity, and (c) the molar heat capacity of the iron cube? Ignore heat losses from the assembly

heat lost by iron=heat gained by water

mcΔT_iron= mcΔT_water

m*c*(70-23 K)=(100 g)*(4.18 J/g*C)*(23-20)

heat capacity = 26.68 J/K

molar heat capacity= heat capacity/molar mass=26.68/55.84= 0.478 J/mol*K

  

13.) When 229 J of energy is supplied as heat to 3.00 mol Ar(g), the temperature of the sample increases by 2.55 K. Calculate the molar heat capacities at constant volume and constant pressure of the gas.

C_P=q/ (n*ΔT)=229/(3mol*2.55K)=29.9 J/mol*K

C_p=C_v+R;  C_v=C_P-R=29.9-8.31=21.6 J/mol*K