Question

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

Answer 1

Explanation:

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

$\huge\mathtt{\fbox{Solution:-}}$

One mole of an ideal gas compressed isothermally and reversibly from a pressure of 2atm to 10atm.

For isothermal process ∆E = 0

and,

$w = - 2.303nRT \: \: log_{10} \frac{ p_{1} }{ p_{2} }$

So,

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

➾$w = + 2.303 \times 600 \times log \: 5$

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

➾$w = + 965.87 \: cal$

➤ For isothermal process,

•.• w = - q

•°• $\boxed{q = - 965.87 cal}$

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