Question

At 27°C, one mole of an ideal gas compressed isothermally and reversibly from a pressure
of 2 atm to 10 atm. Calculate ∆E and q in calorie.
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Answer 1

Answer:

For isothermal process ∆E = 0 and,

$w = - 2.303nrt log_{10} \frac{p1}{p2}$

$w = - 2.303 \times 1 \times 2 \times 300 \times log \frac{2}{10}$

$w = + 2.303 \times 600 \times 0.699$

$w = + 965.87 \: cal$

for isothermal process,

w = -q

q = -965.87 cal

Answer 2

Explanation:

-965 is the correct answer