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Match the process given in column-I with the signs of thermodynamic quantities given in column-II
Column-I Column-II
A Reversible cooling of an ideal gas at
constant volume p $w = 0; q < 0; Δ U < 0$
B Reversible isothermal expansion of ideal gas Q    $w < 0; q > 0; Δ U > 0$
C Adiabatic expansion of non-ideal gas into vacuum R    $w = 0; q = 0; Δ U = 0$
D Reversible melting of sulphur at normal melting point S    $w < 0; q > 0; Δ U = 0$

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$A - P; B - S; C - R; D - Q$

$A - S; B - P; C - Q; D - R$

$A - Q; B - R; C - S; D - P$

$A - R; B - S; C - P; D - Q$

answer is A.

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Detailed Solution

a) $q + w = Δ U$

$Δ v = 0 \Rightarrow w = 0, q = Δ U$

cooling $T_{2} < T_{1} \Rightarrow Δ U < 0$

$w = 0, Δ U < 0, q < 0$

b) $Δ T = 0 \Rightarrow Δ U = 0$

$q + w = Δ U = 0$

$q = - w$

expansion $\Rightarrow w < 0$

$q > 0$

$\Rightarrow w < 0, Δ U = 0, q > 0$

c)Pext = 0 $(∵ V a c u u m)$

w = 0 , adiabatic = q = 0

$q + w = Δ U = 0 + 0$

$w = 0, q = 0, Δ U = 0$

d) melting $\Rightarrow Δ v > 0$

$w < 0, Δ U > 0 \Rightarrow 0$

$q + w = Δ U \Rightarrow q > 0$

So $w < 0, q > 0, Δ U > 0$

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