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JEE MainChemistryModeratePhysical Chemistry

At 27º C, one mole of an ideal gas is compressed isothermally
and reversibly from a pressure of 2 atm to 10 atm. The values of E
and q are (R = 2).

Show answer and solution
W=2.303P2P1 W = 2.303\frac{P_{2}}{P_{1}}\
=2.303×1×2×300log⁡log⁡ 102 =965.84= 2.303 \times 1 \times 2 \times 300\log\log\ \frac{10}{2}\ = 965.84
At constant temperature, E=0E = 0.
E=q+w;q=−w=−965.84 E = q + w;q = - w = - 965.84\ Cal

Source: Grass-G / CH-11 Chemistry Paper-2 1Nov New Jee main.docx

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