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Question: P-V plots for two gases adiabatically are shown in the figure. Plots 1 and 2 should correspond respe...

P-V plots for two gases adiabatically are shown in the figure. Plots 1 and 2 should correspond respectively:

A) HeHe and O2{O_2}
B) O2{O_2} and HeHe
C) HeHe and ArAr
D) O2{O_2} and N2{N_2}

Explanation

Solution

Notice the given PV{{P - V}}plot. Two curves are being plotted. The slope is high for the first curve as compared to the second curve. There are four gases i.e. O2{O_2}, HeHe, ArAr and N2{N_2}. Distinguish the monoatomic, diatomic and triatomic gases from these four. Apply the relation for adiabatic change i.e., PVγ=constant{{P}}{{{V}}^{{\gamma }}}{{ = constant}} and differentiate the relation. After rearranging the terms, a relation between the slope and specific heat can be obtained. After that relate it with each option given in the question.

Complete solution:
Out of O2{O_2}, HeHe, ArAr and N2{N_2}, only O2{O_2} and N2{N_2} are diatomic gases while HeHe and ArAr are monatomic gases.
For diatomic gases, the ratio of specific heats is 1.41.4
For monoatomic gases, the ratio of specific heats is 1.671.67
For adiabatic change, PVγ=constant{{P}}{{{V}}^{{\gamma }}}{{ = constant}}
Taking derivative, we get
dpVγ+γPVγ1dV=0\Rightarrow {{dp}}{{{V}}^{{\gamma }}}{{ + \gamma P}}{{{V}}^{{{\gamma - 1}}}}{{dV = 0}}
On rearranging terms, we get

dpdVVγ+γPVγ1=0 dpdVVγ=γPVγ1\Rightarrow \dfrac{{{{dp}}}}{{{{dV}}}}{{{V}}^{{\gamma }}}{{ + \gamma P}}{{{V}}^{{{\gamma - 1}}}}{{ = 0}} \\\ \therefore \dfrac{{{{dp}}}}{{{{dV}}}}{{{V}}^{{\gamma }}}{{ = - \gamma P}}{{{V}}^{{{\gamma - 1}}}}

Thus, from the above relation we can conclude that slope of the curve varies as per γ{{\gamma }}.
It is clear from the given plot that slope of curve 2 is greater than that of curve 1.
γ2>γ1\therefore {{{\gamma }}_{{2}}}{{ > }}{{{\gamma }}_{{1}}}
Curve 1 should be for diatomic gases like O2{O_2} and N2{N_2}
Curve 2 should be for monoatomic gases like HeHe and ArAr
Thus, curve 1 and curve 2 corresponds to O2{O_2} and HeHe, respectively.

Therefore, option (B) is the correct choice.

Note: In the first option, HeHe and O2{O_2} gases are given. He{{He}} is monoatomic gas while O2{O_2} is diatomic gas. In the second option, O2{O_2} and HeHe gases are given. So, the point is He{{He}} will remain monoatomic gas while O2{{{O}}_{{2}}} will remain diatomic gas but in first option for curve 1, He{{He}} is given and for curve 2, O2{{{O}}_{{2}}} is given. Although the options are the same but the first gas is given for curve 1 and the second gas is for curve 2.