Three moles of hydrogen gas are compressed isothermally and reversibly from 60 L to 20 L and 8.50 kJ of work is done on it. Assuming ideal behavior, calculate the temperature of the gas.

993 K

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1258 K

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1071 K

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none of these

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Solution

The correct option is B 1071 K
For isothermally reversibly Process:-
$w = - n R T ln \frac{V_{2}}{V_{1}}$
or $w = - 2.303 n R T log \frac{V_{2}}{V_{1}} . . . . . . . . (1)$
$V_{1} = 60 L, V_{2} = 20 L, w = 8.50 k J = 8.50 \times 1000 J$
$R =$ Gross constant $= 8.314 J / K m o l$
$n = 3$ ( $3$ moles of $H_{2}$ )
So, from (i) $(8.50 \times 1000) = (- 2) (8.314) (T) log \frac{2}{6}$
$\frac{8500}{- 2 \times 8.314} = T log \frac{1}{3}$
$\frac{8500}{- 2 \times 8.314} = T (- 0.4771)$
$T = \frac{8500}{- 2 \times 8.314 \times 0.4771}$
$T = 1071 K$

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8.50 k J

w = 2.30 n R T l o g \frac{V_{2}}{V_{1}}

n = 3 m o l, V_{1} = 60 L, V_{2} = 20 L, w = 8.5 \times 10^{3} J

8.50 \times 10^{3} J = 2.303 \times (3) (8.314) T

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