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  2. Give answer! - Thermodynamics - JEE Main
# Engineering

 

One mole of an ideal gas passes through a process where pressure and volume obey the

relation P = P_0 [ 1- \frac{1}{2}\left ( \frac{V_0}{V} \right )^2]   .Here P_0  and V_0  are constant . Calculate the change in temperature of

gas if its volume changes from 

V_0 \: \: to \: \: 2 V_0

 

  • Option 1)

     

    \frac{1P_0 V_0 }{2R}

  • Option 2)

    \frac{5 P_0 V_0 }{4 R}

  • Option 3)

    \frac{3 P_0 V_0 }{4 R}

  • Option 4)

    \frac{1 P_0 V_0 }{4 R}

Answers (1)
P Plabita

\frac{nRT }{V} = P_0 [ 1- \frac{1}{2}\left ( \frac{V_0}{V} \right )^2] \\\\ \\T = \frac{P_0 V }{nR} [ 1- \frac{1}{2}\left ( \frac{V_0}{V} \right )^2] \\\\ T_i = \frac{P_0 V_0 }{R} [ 1- \frac{1}{2}\left ( \frac{V_0^2}{V_0^2} \right )].... v = v_0 \\\\ T_f = \frac{P_0 2V_0 }{R} [ 1- \frac{1}{8}\left ( \frac{V_0^2}{V_0^2} \right )]....V = 2 V_0

\delta T = T_f - T_i = \frac{7 P_0 V_0 }{4 R } - \frac{p_0 V_0 }{2 R } = \frac{5 P_0 v_0 }{4 R }


Option 1)

 

\frac{1P_0 V_0 }{2R}

Option 2)

\frac{5 P_0 V_0 }{4 R}

Option 3)

\frac{3 P_0 V_0 }{4 R}

Option 4)

\frac{1 P_0 V_0 }{4 R}

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