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NVAEasyJEE 2023Internal Energy & Enthalpy

JEE Chemistry 2023 Question with Solution

When a 60 W60 \, \text{W} electric heater is immersed in a gas for 100 s100 \, \text{s} in a constant volume container with adiabatic walls, the temperature of the gas rises by 5∘C5^\circ \text{C}. The heat capacity of the given gas is:

Answer

Correct answer:1200

Step-by-step solution

Standard Method

Given: Power of heater = 60 W60 \, \text{W}, time = 100 s100 \, \text{s}, rise in temperature = 5∘C5^\circ \text{C}.

Find: Heat capacity of the gas.

Under adiabatic conditions, the electrical energy supplied by the heater increases the internal energy of the gas at constant volume.

CV×ΔT=P×tC_V \times \Delta T = P \times t

Substituting the values:

CV×5=60×100C_V \times 5 = 60 \times 100

Therefore,

CV=60×1005=1200 J K−1C_V = \frac{60 \times 100}{5} = 1200 \, \text{J K}^{-1}

Therefore, the heat capacity of the gas is 1200 J K−11200 \, \text{J K}^{-1}.

Common mistakes

  • Using the ideal gas equation directly is incorrect here because the required quantity is obtained from energy supplied by the heater, not from PV=nRTPV = nRT. Instead, relate electrical energy PtP t to CVΔTC_V \Delta T.

  • Confusing power with energy is incorrect because 60 W60 \, \text{W} is not the total heat supplied. The total energy added is P×tP \times t, so time must be included before calculating heat capacity.

  • Using heat capacity per mole instead of total heat capacity is incorrect because the question asks for the heat capacity of the given gas sample. Therefore, use CVΔT=PtC_V \Delta T = P t directly for the whole gas.

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