$(a)\;12.56^{\large\circ}C\qquad(b)\;1.256^{\large\circ}C\qquad(c)\;1256.93^{\large\circ}C\qquad(d)\;0.1256^{\large\circ}C$

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Given

P = 50 atm

V = 25 litre

n = 10

$a = 5.46\;atm\;litre^2\;mol^{-2}$

$b = 0.031\;litre\;mol^{-1}$

Now Van der Waal's equation for n mole of gas

$[P + \large\frac{n^2a}{V^2}] [ V - nb] = nRT$

$[50 + \large\frac{100\times5.46}{625}][25-10\times0.031]$

$= 10\times0.0821\times T$

$T = 1529.93K$

$=1256.93^{\large\circ}C$

Hence answer is (C)

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