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Thermodynamics Test - 12

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Thermodynamics Test - 12
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  • Question 1
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    If Vf is the final volume and Vi is the initial volume and pex the external pressure the work done can be calculated by

    Solution

    Work done =- Vi∫Vf pext dVwhere Vi is initial and Vf is final volume. Pex is the external pressure applied on system while change in volume

  • Question 2
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    Enthalpy is defined as

    Solution

    H =U+pV. H is enthalpy of reaction.

  • Question 3
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    Entropy is a state function and measures

    Solution

    Entropy is a state function and it is measure of randomness of a system.

  • Question 4
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    Spontaneity in the context of chemical thermodynamics means

    Solution

    Spontaneity means a reaction occurring on its own without any help of external agency.

  • Question 5
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    Enthalpy of combustion of carbon to CO2is –393.5 kJ mol−1. Calculate the heat released upon formation of 35.2 g of CO2from carbon and dioxygen gas.

    Solution

    when 1 mole of CO2 is produced energy released is –393.5 kJ mol−1 Moles of CO2 given =35.2/44 =0.8 moles So energy released = 0.8 x393.5 KJ/mol = 315 KJ/mol

  • Question 6
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    The volume of gas is reduced to half from its original volume. The specific heat will

    Solution

    Specific heat will remain constant.

  • Question 7
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    A cylinder confines 2.00 L gas under a pressure of 1.00 atm. The external pressure is also 1.00 atm. The gas is heated slowly, with the piston sliding freely outward to maintain the pressure of the gas close to 1.00 atm. Suppose the heating continues until a final volume of 3.50 L is reached. Calculate the work done on the gas.
     

    Solution

  • Question 8
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    A spherical constant temperature heat source of radius r1 is at the center of a uniform solid sphere of radius r2. The rate at which heat is transferred through the surface of the sphere is proportional to

    Solution

    The rate H at which heat is transferred through the slab is,

    (a) directly proportional to the area (A) available.

    (b) inversely proportional to the thickness of the slab Δx.

    (c) directly proportional to the temperature difference ΔT.

    So, H = kA ΔT/ Δx    

    Where k is the proportionality constant and is called thermal conductivity of the material.

    From above we know that the rate H at which heat is transferred through the slab is directly proportional to the area (A) available.

    Area A of the solid sphere is defined as,

    A = 4πr2

    Here r is the radius of the sphere.

    So, the area A1 of a uniform small solid sphere having radius r1 will be,

    A1 = 4πr12

    And, the area A2 of a uniform large solid sphere having radius r2 will be,

    A2 = 4πr22

    Thus the area A from which heat is transferred through the surface of the sphere will be the difference of area of uniform large solid sphere A2 and small solid sphere A1.

    So, A = A2 - A1 = 4πr22 - 4πr12 = 4π (r22 - r12)

    Since the rate H at which heat is transferred through the slab is directly proportional to the area (A) available, therefore the rate at which heat is transferred through the surface of the sphere is proportional to r22 - r12.

  • Question 9
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    For a process from state 1 to state 2, heat transfer in a reversible process is given by

    Solution

    Qreversible = T∆S
    Hence, Q = To(S2-S1)

  • Question 10
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    Suppose a 10.00-kg mass drops through a height difference of 3.00 m, and the resulting work is used to turn a paddle in 200.0 g water, initially at 15.00∘C. The final water temperature is found to be 15.35∘C. Assuming that the work done is used entirely to increase the water temperature, calculate the conversion factor between joules and calories.

    Solution

    1 calorie≈4.2J

  • Question 11
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    The molar heat capacity of a substance is the

    Solution

    The molar heat capacity of a substance is the quantity of heat needed to raise the temperature of one mole by one degree Celsius.

  • Question 12
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    If change in Gibbs energy ΔG is negative (< 0)

    Solution

    ΔG < 0 means reaction is spontaneous. And ΔG >0 means reaction is non spontanenous.

  • Question 13
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    Calculate the standard enthalpy of formation of CH3OH(l) from the following data: 

    CH3​OH(I)+3/2​O2​(g)→CO2​(g)+2H2​O(l);  △r​Hθ=−726kJmol−1

    C(graphite)​+O2​(g)→CO2​(g);    △c​Hθ=−393kJmol−1

    H2​(g)+1/2​O2​(g)→H2​O(l);    △f​Hθ=−286kJmol−1

    Solution

  • Question 14
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    During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is ΔfU0 of formation of CH4 (g) at certain temperature is –393 kJ mol−1. The value of ΔfH0is

    Solution

    C4H10(g) + 13/2O2  →  4CO2(g) + 5H2O(l)
    ∆Ng = (ng)product - (ng)reactant
    = 4-(13/2+1) = -7/2
    or ∆Ng is negative.
    ∆H = ∆U + ∆NgRT
    SInce ∆ng is negative therefore ∆H is less than ∆U.

  • Question 15
    1 / -0

    In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.

    Solution

    for free expansion Pext =0 so w=0 and for adiabatic process q=0 therefore ΔU =0 making ΔT=0

  • Question 16
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    The relationship between Cp and CV for an ideal gas is

    Solution

    For ideal gas Cp - CV = nR. For1 mole of gas n=1.

  • Question 17
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    Standard Molar Enthalpy of Formation is

    Solution

    The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation is enthalpy of formation.

  • Question 18
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    A thermodynamic state function is a physical quantity

    Solution

    State function is one which is only dependent on initial and final state of the system and is independent of the path by which that change has occurred.

  • Question 19
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    Thermodynamics is not concerned about______.

    Solution

    Thermodynamics is not concerned about how and at what rate these energy transformations are carried out, but is based on initial and final states of a system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state.

  • Question 20
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    The pressure-volume work for an ideal gas can be calculated by using the expression   The work can also be calculated from the pV– plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from volume Vi to Vf . choose the correct option.

     

    Solution

    w (reversible) < w (irreversible) (for compression process)

    • Work done is the area under the P−V curve. It can be seen in the curve above that the area under the P−V curve for irreversible compression of the gas is more than the area under the curve for reversible compression.
    • Thus, work done for irreversible compression is more than that for reversible compression.

  • Question 21
    1 / -0

    Suppose that 1.00 kJ of heat is transferred to 2.00 mol argon (at 298 K, 1 atm). What will the final temperature Tf be if the heat is transferred at constant volume?

    Solution

    since volume is constant so w=0 so q=ΔE=nCvΔT Cv=(3/2)R

    Because  argon is monoatomic, approximately ideal gas,

    cv = 3/2 R = 12.47 J K-1 mol-1

    At constant volume,

    qv = n cv Δ T

    1000J = (2.00 mol)(12.47 JK-1mol-1)ΔT

    ΔT = 40.1 k Tf = 298 + 40.1 =338 K

  • Question 22
    1 / -0

    An exothermic process

    Solution

    Exothermic process results in decrease in enthalpy of system because for exothermic reaction ΔH=-ve.

  • Question 23
    1 / -0

    For the process to occur under adiabatic conditions, the correct condition is:

    Solution

    For adiabatic process Δq=0.

  • Question 24
    1 / -0

    For an isolated system, ΔU = 0, what will be ΔS?

    Solution

    For an isolated system, ∆U = 0 and for a spontaneous process, total entropy change must be positive.

     For example, consider the diffusion of two gases A and B into each other in a closed container which is isolated from the surroundings.
    The two gases A and B are separated by a movable partition. When partition is removed, the gases begin to diffuse into each other and the system becomes more disordered. It shows that ∆S > 0 and ∆U = 0 for this process.

     

  • Question 25
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    The entropy change can be calculated by using the expression When water freezes in a glass beaker, choose the correct statement amongst the following :

    Solution

    For freezing of process since process is spontaneous therefore if ΔS (system) decreases but ΔS (surroundings) increases.

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