∵dH=dq (at P= constant) dH=CpdTdT(dTdH)P=CPdU=dq (at V= constant) dU=CVdT(dTdU)V=CV∵dG=VdP−SdT at P= constant, dP=0 (dTdG)P=−S at T= constant (dPdG)T=V
Match List - I with List - II. 
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Held on 22 Jan 2025 · Verified 6 Jul 2026.
(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
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The ratio of mass percentage (w/w) of C : H in a hydrocarbon is $12 : 1$. It has two carbon atoms. The weight (in g) of $CO_2(g)$ formed when $3.38$ g of this hydrocarbon is completely burnt in oxygen is : (Given : Molar mass in g mol$^{-1}$ C : 12, H : 1, O : 16)
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For the given reaction:<br>$\mathrm{CaCO}_{3}+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$<br>If $90 \mathrm{~g} \mathrm{CaCO}_{3}$ is added to 300 mL of HCl which contains $38.55 \% \mathrm{HCl}$ by mass and has density $1.13 \mathrm{~g} \mathrm{~mL}^{-1}$, then which of the following option is correct ?<br>Given molar mass of $\mathrm{H}, \mathrm{Cl}, \mathrm{Ca}$ and O are 1, 35.5, 40 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively.
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