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CBSE-XI-Physics

27: Specific Heat Capacities of Gases

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  • #
    Section : i
  • Qstn #1
    Does a gas have just two specific heat capacities or more than two? Is the number of specific heat capacities of a gas countable?
    Ans : No, a gas doesn't have just two specific heat capacities, as the heat capacities depend on the process followed. There are infinite processes; therefore, there can be infinite number of specific heat capacities.
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  • Qstn #2
    Can we define specific heat capacity at constant temperature?
    Ans : Specific heat capacity, `` s=\frac{\,\mathrm{\,\Delta \,}Q}{m\,\mathrm{\,\Delta \,}T}``, where `` \Delta ``Q/m is the heat supplied per unit mass of the substance and `` \,\mathrm{\,\Delta \,}``T is the change in temperature produced. At constant temperature, ​`` \,\mathrm{\,\Delta \,}``T = 0; therefore, s = infinity. So, we cannot define specific heat capacity at constant temperature.
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  • Qstn #3
    Can we define specific heat capacity for an adiabatic process?
    Ans : Specific heat capacity, `` s=\frac{\,\mathrm{\,\Delta \,}Q}{m\,\mathrm{\,\Delta \,}T}``, where`` ∆``Q/m is the heat supplied per unit mass of the substance and `` ∆``T is the change in temperature produced. In an adiabatic process, no heat exchange is allowed; so, `` ∆``Q = 0 and hence, s = 0. Therefore, in an adiabatic process, specific heat capacity is zero.
    Page No 76:
  • Qstn #4
    Does a solid also have two kinds of molar heat capacities Cp and Cv? If yes, is Cp > Cv? Or is Cp - Cv = R?
    Ans : Yes, a solid also has two kinds of molar heat capacities, Cp and Cv. In a solid, expansion coefficient is quite small; therefore dependence of heat capacity on the process is negligible. So, Cp > Cv with just a small difference, which is not equal to R.
    Page No 76:
  • Qstn #5
    In a real gas, the internal energy depends on temperature and also on volume. The energy increases when the gas expands isothermally. Examining the derivation of Cp - Cv = R, find whether Cp - Cv will be more than R, less than R or equal to R for a real gas.
    Ans : In a real gas, as the internal energy depends on temperature and volume, the derived equation for an ideal gas `` (\,\mathrm{\,d\,}Q{)}_{\,\mathrm{\,p\,}}=(\,\mathrm{\,d\,}Q{)}_{\,\mathrm{\,v\,}}+nR\,\mathrm{\,d\,}T`` will change to ​`` (\,\mathrm{\,d\,}Q{)}_{\,\mathrm{\,p\,}}=(\,\mathrm{\,d\,}Q{)}_{\,\mathrm{\,v\,}}+nR\,\mathrm{\,d\,}T`` + k, where k is the change in internal energy (positive) due to change in volume when pressure is kept constant. So, in the case of a real gas, for n=1 mole (say),
    `` {C}_{\,\mathrm{\,p\,}}-{C}_{\,\mathrm{\,v\,}}=R+\frac{k}{\,\mathrm{\,d\,}T}``
    `` \Rightarrow {C}_{\,\mathrm{\,p\,}}-{C}_{\,\mathrm{\,v\,}}>R,``
    where Cp and Cv are the specific heat capacities at constant pressure and volume, respectively.
    Page No 76:
  • Qstn #6
    Can a process on an ideal gas be both adiabatic and isothermal?
    Ans : According to the first law of thermodynamics, change in internal energy, `` ∆``U is equal to the difference between heat supplied to the gas, `` ∆``Q and the work done on the gas,​`` ∆``W, such that `` \,\mathrm{\,\Delta \,}Q=\,\mathrm{\,\Delta \,}U+\,\mathrm{\,\Delta \,}W``​. In an adiabatic process, `` ∆``Q = 0 and in an isothermal process, change in temperature, `` ∆``T = 0. Therefore,
    `` \,\mathrm{\,\Delta \,}Q=\,\mathrm{\,\Delta \,}U+\,\mathrm{\,\Delta \,}W``
    `` \Rightarrow \,\mathrm{\,\Delta \,}Q=n{C}_{v}\,\mathrm{\,\Delta \,}T+\,\mathrm{\,\Delta \,}W``
    `` \Rightarrow 0=n{C}_{v}\left(0\right)+\,\mathrm{\,\Delta \,}W``
    `` \Rightarrow \,\mathrm{\,\Delta \,}W=0,``
    where Cv is the heat capacity at constant volume.
    This shows that if the process is adiabatic as well as isothermal, no work will be done. So, a process on an ideal gas cannot be both adiabatic and isothermal.
    Page No 76:
  • Qstn #7
    Show that the slope of the p-V diagram is greater for an adiabatic process compared to an isothermal process.
    Ans : In an isothermal process,
    PV = k ...(i)
    On differentiating it w.r.t V, we get
    `` V\frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}+P=0``
    `` \frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}=-\frac{P}{V}``
    `` \frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}=-\frac{k}{{V}^{2}}[\text{U}\,\mathrm{\,sing\,}(\,\mathrm{\,i\,}\left)\right],\text{k = constant}``
    k = constant
    In an adiabatic process,
    `` P{V}^{\gamma }=K....\left(\,\mathrm{\,ii\,}\right)``
    On differentiating it w.r.t V, we get
    `` {V}^{\gamma }\frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}+\gamma P{V}^{\gamma -1}=0``
    `` \frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}=-\frac{\gamma P{V}^{\gamma -1}}{{V}^{\gamma }}``
    `` \frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}=-\frac{\gamma K}{{V}^{\gamma +1}}[\text{U}\,\mathrm{\,sing\,}(\,\mathrm{\,ii\,}\left)\right],\gamma >1\text{and}``
    K is constant
    `` \gamma \text{and}\frac{\,\mathrm{\,d\,}P}{\,\mathrm{\,d\,}V}`` are the slope of the curve and the ratio of heat capacities at constant pressure and volume, respectively; P is pressure and V is volume of the system.
    By comparing the two slopes and keeping in mind that `` \gamma >1``, we can see that the slope of the P-V diagram is greater for an adiabatic process than an isothermal process.
    Page No 76:
  • Qstn #8
    Is a slow process always isothermal? Is a quick process always adiabatic?
    Ans : For an isothermal process, PV = K, where P is pressure, V is volume of the system and K is constant. In an isothermal process, a small change in V produces only a small change in p, so as to keep the product constant. On the other hand, in an adiabatic process, `` P{V}^{\gamma }=k,\gamma =\frac{{C}_{\,\mathrm{\,P\,}}}{{C}_{\,\mathrm{\,V\,}}}`` > 1 is the ratio of heat capacities at constant pressure and volume, respectively, and k is a constant. In this process, a small increase in volume produces a large decrease in pressure. Therefore, an isothermal process is considered to be a slow process and an adiabatic process a quick process.
    Page No 76:
  • Qstn #9
    Can two states of an ideal gas be connected by an isothermal process as well as an adiabatic process?
    Ans : For two states to be connected by an isothermal process,
    `` {P}_{1}{V}_{1}={P}_{2}{V}_{2}`` ...(i)
    For the same two states to be connected by an adiabatic process,
    `` {P}_{1}{{V}_{1}}^{\gamma }={P}_{2}{{V}_{2}}^{\gamma }`` ...(ii)
    If both the equations hold simultaneously then, on dividing eqaution (ii) by (i) we get
    `` {{V}_{1}}^{\gamma -1}={{V}_{2}}^{\gamma -1}``
    Let the gas be monatomic. Then,
    `` \gamma =\frac{5}{3}``
    `` \Rightarrow {{V}_{1}}^{\frac{2}{3}}={{V}_{2}}^{\frac{2}{3}}``
    `` \Rightarrow {V}_{1}={V}_{2}``
    If this condition is met, then the two states can be connected by an isothermal as well as an adiabatic process.
    Page No 76:
  • Qstn #10
    The ratio Cp / Cv for a gas is 1.29. What is the degree of freedom of the molecules of this gas?
    Ans : For the molecules of a gas, `` \gamma =\frac{{C}_{\,\mathrm{\,p\,}}}{{C}_{\,\mathrm{\,v\,}}}=1+\frac{2}{f}``,
    where f is the degree of freedom.
    `` \,\mathrm{\,Given\,}:\gamma =1.29``
    `` \Rightarrow 1+\frac{2}{f}=1.29=\frac{9}{7}``
    `` \Rightarrow \frac{2}{f}=\frac{9}{7}-1``
    `` \Rightarrow \frac{2}{f}=\frac{2}{7}``
    `` \Rightarrow f=7``
    `` ``
    `` ``
    Therefore, the molecules of this gas have 7 degrees of freedom.
    But in reality, no gas can have more than 6 degrees of freedom.
    Page No 76:
  • #
    Section : ii
  • Qstn #1
    Work done by a sample of an ideal gas in a process A is double the work done in another process B. The temperature rises through the same amount in the two processes. If CA and CB be the molar heat capacities for the two processes,
    (a) CA = CB
    (b) CA < CB
    (c) CA > CB
    (d) CA and CB cannot be defined.
    digAnsr:   c
    Ans : (c) CA > CB
    According to the first law of thermodynamics, `` \,\mathrm{\,\Delta \,}Q=\,\mathrm{\,\Delta \,}U+\,\mathrm{\,\Delta \,}W``, where `` ∆``Q is the heat supplied to the system when `` \Delta ``W work is done on the system and `` ∆``U is the change in internal energy produced. Since the temperature rises by the same amount in both processes, change in internal energies are same, i.e. `` ∆``UA = `` ∆``UB.
    But as, `` ∆``WA = `` ∆``WB this gives `` \Delta ``QA = 2`` \Delta ``QB.
    Now, molar heat capacity of a gas, `` C=\frac{\,\mathrm{\,\Delta \,}Q}{n\,\mathrm{\,\Delta \,}T}``, where `` ∆``Q/n is the heat supplied to a mole of gas and `` ∆``T is the change in temperature produced. As `` ∆``QA = 2`` ∆``QB, CA > CB.
    Page No 76:
  • Qstn #2
    For a solid with a small expansion coefficient,
    (a) Cp - Cv = R
    (b) Cp = Cv
    (c) Cp is slightly greater than Cv
    (d) Cp is slightly less than Cv
    Ans : c) Cp is slightly greater than Cv
    For a solid with a small expansion coefficient, work done in a process will also be small. Thus, the specific heat depends slightly on the process. Therefore, Cp is slightly greater than Cv.
    Page No 76:
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