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  • Two different adiabatic parts for the same gas intersects two isotherms at <img alt="\mathrm{T_1 \: and\: T_2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BT_1%20%5C%3A%20and%5

Two different adiabatic parts for the same gas intersects two isotherms at \mathrm{T_1 \: and\: T_2} as shown in the P-V diagram in the figure. The ratio \mathrm{\frac{V_a}{V_d}} compare with the ratio of \mathrm{\frac{V_b}{V_c}} is:

Option: 1

Less than 1


Option: 2

Equal to 1
 


Option: 3

Greater than 1


Option: 4

Either (A) or (B)


Answers (1)

For adiabatic change

\mathrm{ \mathrm{PV}^\gamma=\text { constant } }

\mathrm{\text { since } \mathrm{PV}=\mathrm{nRT} }

\mathrm{\therefore \mathrm{TV}^{\gamma-1}=\text { constant } }

For adiabatic process,\mathrm{ BC }

\mathrm{ \mathrm{T}_1 \mathrm{~V}_{\mathrm{b}}^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_{\mathrm{c}}^{\gamma-1} }          (i)

and for adiabatic process \mathrm{ DA, }

\mathrm{ \mathrm{T}_1 \mathrm{~V}_{\mathrm{a}}{ }^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_{\mathrm{d}}{ }^{\gamma-1} }       (ii)

Dividing equation (ii) by (i),

\mathrm{ \left(\frac{V_a}{V_b}\right)^{\gamma-1}=\left(\frac{V_d}{V_c}\right)^{\gamma-1} }

\mathrm{ \text { or } \quad \left(\frac{V_a}{V_d}\right)=\left(\frac{V_b}{V_c}\right)}

Hence, both ratios are same.

Posted by

Sumit Saini

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