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Q.

5 mol of $N_{2} (g)$ at 5 atmospheric pressure contained in a 100L cylinder absorbed 30.26kJ of heat when it expanded to 200 L at 2 atmospheric pressure. The change in the internal energy of $N_{2}$ gas is:

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a

10kJ

b

-20.26 kJ

c

+20.26 J

d

+20.26 kJ

answer is D.

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Detailed Solution

$V_{1} = 100 L; V_{2} = 200 L; P = 2 atm$

Work, $W = - P (V_{2} - V_{1}) = - 2 atm (200 - 100) L = - 200 L atm = - 200 \times 101.30 J [∵ 1 Latm = 101.30 J]$

Heat absorbed, q=30.26 kJ

$But ΔU = q + W (first law of thermodynamics)$

$∴ ΔU = 30.26 kJ + (- 20.26 kJ) = + 10 kJ$

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5 mol of N2(g) at 5 atmospheric pressure contained in a 100L cylinder absorbed 30.26kJ of heat when it expanded to 200 L at 2 atmospheric pressure. The change in the internal energy of N2 gas is: