A gas (Molar mass = 280 ~g ~mol^-1 ) was burnt in excess O_2 in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 ~K to 298.45K . If the heat capacity of calorimeter is 2.5 ~kJ ~K^-1 and enthalpy of combustion of gas is 9 ~kJ ~mol^-1 then amount of gas burnt is \underline\hspace2cm g. (Nearest Integer)

Thermodynamics PYQ — JEE Main 2022 · 27 Jul · Shift 2 · Q16 with Solution

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Correct answer: 35

Molar mass of gas $= 280\,\mathrm{g\,mol^{-1}}$
Temperature rise of calorimeter: $\Delta T = 298.45 - 298.0 = 0.45\,\mathrm{K}$
Heat capacity of calorimeter: $C_{\text{cal}} = 2.5\,\mathrm{kJ\,K^{-1}}$
Enthalpy of combustion of gas: $\Delta H_{\text{comb}} = 9\,\mathrm{kJ\,mol^{-1}}$

Since the calorimeter temperature increases, heat released by combustion is absorbed by the calorimeter: $q_{\text{cal}} = C_{\text{cal}}\Delta T$

So, $q_{\text{cal}} = 2.5 \times 0.45 = 1.125\,\mathrm{kJ}$

Thus, heat released by the burnt gas is: $q = 1.125\,\mathrm{kJ}$

Given enthalpy of combustion is $9\,\mathrm{kJ}$ per mole, so if $n$ moles are burnt: $n \times 9 = 1.125$

Hence, $n = \frac{1.125}{9} = 0.125\,\mathrm{mol}$

$m = nM = 0.125 \times 280 = 35\,\mathrm{g}$

$\boxed{35}$