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Thermodynamics Test - 77

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Thermodynamics Test - 77
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  • Question 1
    1 / -0
     A gas mixture consists of $$2$$ moles of O, and $$4$$ moles of Ar at a temperature T. Neglecting all vibrational moles, the total internal energy of the system is 
    Solution

  • Question 2
    1 / -0
    Hess's law states that:
    Solution

  • Question 3
    1 / -0
    Consider $$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$$ If the standard emf is $$E_{cell}^o=2.0V$$ and $$F=96500C$$
    Find $$\Delta G^0$$(kJmol)
    Solution
    Solution:- (A) $$-388$$
    $$\Delta G^0=-nF E_{Cell}^0=-2\times 96500 \times 2.0\times 10^{-3}=-388kJ/mol$$
  • Question 4
    1 / -0
    Assertion : A reaction which is spontaneous and accompanied by decrease of randomness must be exothermic.
    Reason : All exothermic reactions are accompanied by decrease of randomness.
  • Question 5
    1 / -0
    In which case, process will be spontaneous at all temperatures?
    Solution
    Solution:- (A) $$\Delta{H} < 0$$ and $$\Delta{S} > 0$$
    For spontaneous process $$\Delta G=\Delta H-T\Delta S$$
    $$\because \Delta H<0$$
    $$\Delta S>0$$
  • Question 6
    1 / -0
    The heat energy of $$743J$$ is needed to raise the temperature of $$5$$ moles of an ideal gas by $$2K$$ at constant pressure. How much heat energy is needed to raise the temperature of the same mass of the gas by $$2K$$ at constant volume?
    Solution
    as we know,

    $$(Q)_P=nC_PΔT=\Delta H$$ 

    $$C_P−C_V=R⇒C_V=C_P−R,\ (Q)_V=nC_VΔT=\Delta U$$

    At constant pressure,

    $$\Delta H=nC_P​ΔT=743J$$

    At constant volume,

    $$\Delta U=nC_v​ΔT=n(C_p​−R)\Delta T=nC_P​ΔT-nRΔT$$

    $$\Delta U=743−(5×8.3×2)$$

    $$\Delta U=660J$$

    Option C is correct.
  • Question 7
    1 / -0

    A student mixed 25.0cm3 of 4.00 moldm–3 hydrochloric acid with an equal volume of 4.00moldm–3 sodium hydroxide.?

    The initial temperature of both solutions was 15.0°C. The
    maximum temperature recorded was 30.0°C.
    Using these results, what is the enthalpy change of neutralisation of hydrochloric acid?

    Solution
    Calculate heat absorbed by the water:
    (Assume density of solutions = 1.00 g/cm^3 and specific heat of solution is the same as water)
    q = m c (T2-T1)
    q = 50 g (4.184 J/gC) ( 30-15 C) = 3138 J

    Heat released by neutralization reaction = -3138 J
    Moles HCl used = 0.0250 L X 4.00 mol/L = 0.100 mol

    Delta H = -3138 J / 0.100 mol = -3.14X10^4 J = -31.4 kJ/mol
    B is the correct answer.
  • Question 8
    1 / -0
    For a system in equilibrium, $$\Delta G=0$$ under conditions of constant:
    Solution
    For a system in equilibrium $$\Delta G=0$$at constant pressure and temperature.

    $$G(p,T)=U+PV-TS$$

    $$G(p,T)=H-TS$$

    So, Gibbs's free energy is defined for constant pressure and temperature.

    Hence, the correct option is A
  • Question 9
    1 / -0

    Directions For Questions

    One mole of an ideal gas whose adiabatic exponent equals $$\gamma$$ undergoes a process in which the gas pressure relates to the temperature as $$p=a {T}^{\alpha }$$, where $$a$$ and $$\alpha $$ are constants

    ...view full instructions

    At what value of $$\alpha$$ will the heat capacity be negative?
  • Question 10
    1 / -0
    What is the heat capacity of the ideal monoatomic gas undergoing the process shown in the figure?

    Solution

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