JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Chemical Thermodynamics

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 298.45 K. 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

____

g. (Nearest Integer)

Answer: 35

Solution

Heat released = Heat capacity × Temperature change
Heat released = 2.5 × (298.45 - 298.0) = 2.5 × 0.45 = 1.125 kJ

For constant volume, ΔH ≈ ΔU (since Δn_g is small)

If x grams of gas is burnt:
Moles =

\frac{x}{280}

Heat released by

\frac{x}{280}

moles = 1.125 kJ

Heat released by 1 mole =

\frac{1.125 \times 280}{x}

kJ = 9 kJ (given)

Solving:

\frac{1.125 \times 280}{x} = 9

x = \frac{1.125 \times 280}{9} = \frac{315}{9} = 35

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Chemical Thermodynamics

Solution

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