A) 2 : 1 : 1
B) 1 : 1 : 1
C) 1 : 2 : 1
D) 1 : 1 : 2
Correct Answer: B
Solution :
Given: \[{{V}_{2}}=2V,\]and\[\gamma =3/2\] For the case of adiabatic process and for sample X, we have \[P{{V}^{\gamma }}=cons\tan t\] Hence, \[{{P}_{1}}V_{1}^{\gamma }={{P}_{2}}V_{2}^{\gamma }\] \[{{({{P}_{2}})}_{x}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma }}{{P}_{1}}={{\left( \frac{{{V}_{1}}}{2{{V}_{1}}} \right)}^{\gamma }}{{P}_{1}}\] \[={{\left( \frac{1}{2} \right)}^{3/2}}{{P}_{1}}=\frac{({{P}_{1}})X}{2\sqrt{2}}\] For the case of isobaric process in which pressure remains constant and so, for sample Y \[{{({{P}_{2}})}_{\gamma }}={{({{P}_{1}})}_{\gamma }}\] ?(ii) For the case of isothermal process and for sample Z PV =RT= constant Hence, \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\] \[{{({{P}_{2}})}_{z}}=\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right){{P}_{1}}=\left( \frac{{{V}_{1}}}{2{{V}_{1}}} \right){{P}_{1}}=\frac{{{({{P}_{1}})}_{z}}}{2}\] ?(iii) Let the ratio of initial pressure according X, Y, Z sample be \[2\sqrt{2}:1:2=k\] \[\therefore \] \[{{({{P}_{1}})}_{X}}=2\sqrt{2}k\]and \[{{({{P}_{1}})}_{\gamma }}=k\] and \[{{({{P}_{1}})}_{z}}=2k\] Now, putting the values of\[{{({{P}_{1}})}_{X}},{{({{P}_{1}})}_{Y}},({{P}_{1}})Z\] and \[{{({{P}_{1}})}_{2}}\]in Eqs. (i), (ii) and (iii) respectively, wg get \[{{({{P}_{2}})}_{X}}=\frac{2\sqrt{2}k}{2\sqrt{2}}=k\] ?(iv) \[{{({{P}_{2}})}_{Y}}=k\] ?(v) \[{{({{P}_{2}})}_{Z}}=\frac{2k}{2}=k\] ?(vi) Therefore, from above Eqs. (iv), (v) and (vi) it is quite clear them \[{{({{P}_{2}})}_{X}}:{{({{P}_{2}})}_{\gamma }}:{{({{P}_{2}})}_{Z}}=k:k:k=1:1:1\]
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