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NVAEasyJEE 2025Internal Energy & Enthalpy

JEE Chemistry 2025 Question with Solution

A perfect gas (0.1 mol0.1 \, \text{mol}) having CˉV=1.50 R\bar{C}_V = 1.50\,R (independent of temperature) undergoes the above transformation from point 11 to point 44. If each step is reversible, the total work done (ww) while going from point 11 to point 44 is (\hspace{3cm) J (nearest integer) [Given : R=0.082 L atm K−1R = 0.082 \, \text{L atm K}^{-1}]

P-V diagram with pressure on vertical axis in atm and volume on horizontal axis in cm cubed, showing points 1, 2, 3, 4 forming a rectangle with pressures 1.00 and 3.00 atm and volumes 1000 and 2000 cm cubed, arrows indicating path 1 to 2 to 3 to 4.

Answer

Correct answer:304

Step-by-step solution

Standard Method

Given: The gas goes from point 11 to point 44 through three reversible steps on the PP-VV diagram.

  • 1→21 \to 2 is isochoric at V=1000 cm3V = 1000 \, \text{cm}^3
  • 2→32 \to 3 is isobaric at P=3.00 atmP = 3.00 \, \text{atm}
  • 3→43 \to 4 is isochoric at V=2000 cm3V = 2000 \, \text{cm}^3

Find: Total work done ww from point 11 to point 44.

For a reversible process,

w=−∫P dVw = - \int P \, dV

So the total work is

wtotal=w1→2+w2→3+w3→4w_{\text{total}} = w_{1 \to 2} + w_{2 \to 3} + w_{3 \to 4}

For the isochoric steps, dV=0dV = 0, so the work done is zero.

w1→2=0w_{1 \to 2} = 0 w3→4=0w_{3 \to 4} = 0

For the isobaric step 2→32 \to 3,

w2→3=−PΔVw_{2 \to 3} = -P\Delta V

Here,

ΔV=2000−1000=1000 cm3=1 L\Delta V = 2000 - 1000 = 1000 \, \text{cm}^3 = 1 \, \text{L}

Thus,

w2→3=−(3.00 atm)(1 L)=−3.00 L atmw_{2 \to 3} = -(3.00 \, \text{atm})(1 \, \text{L}) = -3.00 \, \text{L atm}

Now add the contributions:

wtotal=0+(−3.00 L atm)+0=−3.00 L atmw_{\text{total}} = 0 + (-3.00 \, \text{L atm}) + 0 = -3.00 \, \text{L atm}

Convert to joules using

1 L atm=101.325 J1 \, \text{L atm} = 101.325 \, \text{J}

So,

wtotal=−3.00×101.325=−303.975 Jw_{\text{total}} = -3.00 \times 101.325 = -303.975 \, \text{J}

Rounding to the nearest integer,

wtotal≈−304 Jw_{\text{total}} \approx -304 \, \text{J}

The blank asks for the numerical value, so the required answer is 304304.

Step-by-step Work Analysis

Given: Only the horizontal segment of the path changes volume. Vertical segments have constant volume.

Find: Numerical value of the total work done.

  1. In the path 1→21 \to 2, volume remains constant at 1000 cm31000 \, \text{cm}^3. Therefore,
ΔV=0⇒w=0\Delta V = 0 \Rightarrow w = 0
  1. In the path 2→32 \to 3, pressure is constant at 3.00 atm3.00 \, \text{atm} and the gas expands from 1000 cm31000 \, \text{cm}^3 to 2000 cm32000 \, \text{cm}^3.
ΔV=1000 cm3=1 L\Delta V = 1000 \, \text{cm}^3 = 1 \, \text{L}

Hence,

w=−PΔV=−(3.00)(1)=−3.00 L atmw = -P\Delta V = -(3.00)(1) = -3.00 \, \text{L atm}
  1. In the path 3→43 \to 4, volume again remains constant, so
w=0w = 0

Therefore,

wtotal=−3.00 L atmw_{\text{total}} = -3.00 \, \text{L atm}

Using the conversion,

−3.00 L atm=−3.00×101.325=−303.975 J-3.00 \, \text{L atm} = -3.00 \times 101.325 = -303.975 \, \text{J}

So the magnitude to be filled is 304304. The final answer is 304304.

Common mistakes

  • Treating the vertical segments 1→21 \to 2 and 3→43 \to 4 as if they do work. This is wrong because in an isochoric process dV=0dV = 0, so w=−∫P dV=0w = -\int P \, dV = 0. Only segments with volume change contribute to work.

  • Using the wrong sign convention for chemistry. For expansion, chemistry uses w=−PΔVw = -P\Delta V, so the work here is negative. If the question asks for the number to fill in the blank, use the required numerical value after interpreting the sign from the statement.

  • Forgetting to convert 1000 cm31000 \, \text{cm}^3 into 1 L1 \, \text{L} before using L atm\text{L atm} units. Pressure is given in atm, so the volume change must be in liters to compute work correctly.

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