$1 m o l e$ of a gas is changed from its initial state $(15 l i t; 2 a t m)$ to final state $(4 l i t, 10 a t m)$ reversinly. If this change can be represented by a straight line in $p - V$ curve, calculate maximum temperature, the gas attained.

Question

Answer 1

$(V_{1}, P_{1}) = (15 l i t, 2 a t m)$
$(V_{2}, P_{2}) = (4 l i t, 10 a t m)$

Eq. of line
$P - P_{1} = \frac{P _{1} - P _{2}}{V _{1} - V _{2}} (V - V_{1})$

$P = P_{1} + \frac{P _{1} - P _{2}}{V _{1} - V _{2}} (V - V_{1})$

$\frac{n R T}{V} = P_{1} + (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V - (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V_{1}$

$T = \frac{1}{n R} {P_{1} V + (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V^{2} - (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V_{1} . V}$

For $T$ to be max $\frac{d T}{d V} = 0$

$\frac{d T}{d V} = \frac{1}{n R} {P_{1} + 2 (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V - (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V_{1}} = 0$

$2 (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V = \frac{P _{1} V _{1} - P _{2} V _{1}}{V _{1} - V _{2}} - P_{1} = \frac{P _{1} V _{1} - P _{2} V _{1} - P _{1} V _{1} + P _{1} V _{2}}{V _{1} - V _{2}} = \frac{P _{1} V _{2} - P _{2} V _{1}}{( V _{1} - V _{2} )}$

$V = \frac{P _{1} V _{2} - P _{2} V _{1}}{2 ( P _{1} - P _{2} )} = \frac{2 \times 4 - 1 0 \times 1 5}{2 ( 2 - 1 0 )} = 8.875$

$T_{m a x}$ , for $1 m o l e$ , $n = 1$

$T_{m a x} = \frac{1}{R} {2 \times 8.875 + (\frac{2 - 1 0}{1 5 - 4}) \times (8.875)^{2} - (\frac{2 - 1 0}{1 5 - 4}) \times 15 \times 8.875}$

$= \frac{1}{R} {17.75 + (\frac{- 8}{1 1}) \times 78.7656 - (\frac{- 8}{1 1}) \times 133.125}$

$= \frac{1}{R} {17.75 + 96.82 + (- 57.28)} = \frac{1}{R} \times 57.286 = \frac{5 7 . 2 9}{0 . 0 8 2 1} = 697.80 K$

Answer 2

(V_{1}, P_{1}) = (15 l i t, 2 a t m)

(V_{2}, P_{2}) = (4 l i t, 10 a t m)

Eq. of line

P - P_{1} = \frac{P _{1} - P _{2}}{V _{1} - V _{2}} (V - V_{1})

P = P_{1} + \frac{P _{1} - P _{2}}{V _{1} - V _{2}} (V - V_{1})

\frac{n R T}{V} = P_{1} + (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V - (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V_{1}

T = \frac{1}{n R} {P_{1} V + (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V^{2} - (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V_{1} . V}

For

T

to be max

\frac{d T}{d V} = 0

\frac{d T}{d V} = \frac{1}{n R} {P_{1} + 2 (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V - (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V_{1}} = 0

2 (\frac{P _{1} - P _{2}}{V _{1} - V _{2}}) V = \frac{P _{1} V _{1} - P _{2} V _{1}}{V _{1} - V _{2}} - P_{1} = \frac{P _{1} V _{1} - P _{2} V _{1} - P _{1} V _{1} + P _{1} V _{2}}{V _{1} - V _{2}} = \frac{P _{1} V _{2} - P _{2} V _{1}}{( V _{1} - V _{2} )}

V = \frac{P _{1} V _{2} - P _{2} V _{1}}{2 ( P _{1} - P _{2} )} = \frac{2 \times 4 - 1 0 \times 1 5}{2 ( 2 - 1 0 )} = 8.875

T_{m a x}

, for

1 m o l e

,

n = 1

T_{m a x} = \frac{1}{R} {2 \times 8.875 + (\frac{2 - 1 0}{1 5 - 4}) \times (8.875)^{2} - (\frac{2 - 1 0}{1 5 - 4}) \times 15 \times 8.875}

= \frac{1}{R} {17.75 + (\frac{- 8}{1 1}) \times 78.7656 - (\frac{- 8}{1 1}) \times 133.125}

= \frac{1}{R} {17.75 + 96.82 + (- 57.28)} = \frac{1}{R} \times 57.286 = \frac{5 7 . 2 9}{0 . 0 8 2 1} = 697.80 K