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NVAEasyJEE 2023Internal Energy & Enthalpy

JEE Chemistry 2023 Question with Solution

1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27∘C27^\circ \text{C}. The work done is 3 kJ mol−13 \, \text{kJ mol}^{-1}. The final temperature of the gas is ..... K\text{K} (Nearest integer). Given Cv=20 J mol−1 K−1C_v = 20 \, \text{J mol}^{-1} \, \text{K}^{-1}.

Answer

Correct answer:150

Step-by-step solution

Standard Method

Given: 11 mole of an ideal gas expands reversibly and adiabatically. Initial temperature is 300 K300 \, \text{K}, work done is 3 kJ mol−1=3000 J mol−13 \, \text{kJ mol}^{-1} = 3000 \, \text{J mol}^{-1}, and Cv=20 J mol−1 K−1C_v = 20 \, \text{J mol}^{-1} \, \text{K}^{-1}.

Find: The final temperature T2T_2.

For an adiabatic process, q=0q = 0, so ΔU=w\Delta U = w.

Using the relation:

w=CvΔTw = C_v \Delta T

Substitute the given values:

1×20×(T2−300)=−30001 \times 20 \times (T_2 - 300) = -3000

Now solve:

T2−300=−150T_2 - 300 = -150 T2=150 KT_2 = 150 \, \text{K}

Therefore, the final temperature of the gas is 150 K150 \, \text{K}.

Energy Change Interpretation

Given: The process is adiabatic, so no heat is exchanged. The gas does work during expansion.

Find: The nearest integer value of the final temperature.

Since the gas expands adiabatically, its internal energy decreases by the amount of work done. Therefore:

ΔU=nCv(T2−T1)=−3000 J mol−1\Delta U = n C_v (T_2 - T_1) = -3000 \, \text{J mol}^{-1}

With n=1n = 1, Cv=20 J mol−1 K−1C_v = 20 \, \text{J mol}^{-1} \, \text{K}^{-1}, and T1=27∘C=300 KT_1 = 27^\circ \text{C} = 300 \, \text{K}:

1×20×(T2−300)=−30001 \times 20 \times (T_2 - 300) = -3000

Divide by 2020:

T2−300=−150T_2 - 300 = -150

Hence,

T2=150 KT_2 = 150 \, \text{K}

So, the required numerical answer is 150.

Common mistakes

  • Using 2727 directly instead of converting 27∘C27^\circ \text{C} to 300 K300 \, \text{K} is incorrect. Temperature change equations here must use Kelvin. Convert first, then substitute.

  • Taking the sign of work incorrectly is a common mistake. During adiabatic expansion, the gas does work and its internal energy decreases, so T2−T1T_2 - T_1 must be negative here.

  • Using CpC_p instead of CvC_v is wrong because the internal energy change of an ideal gas is written as ΔU=nCvΔT\Delta U = n C_v \Delta T. Use the given CvC_v value.

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