Q

# Give answer! - Thermodynamics - JEE Main

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)

$\frac{1 P_0 V_0 }{4 R}$

Views

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

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

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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