One mole of an ideal gas expands isothermally and reversibly from 10 to 20 at 300 K. , and work done in the process respectively are [2025]

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Solution

Q.
One mole of an ideal gas expands isothermally and reversibly from 10 ${dm}^{3}$ to 20 ${dm}^{3}$ at 300 K. $ΔU$ , $q$ and work done in the process respectively are:

Given: R = 8.3 J $K^{- 1}$ ${mol}^{- 1}$

ln 10 = 2.3
log 2 = 0.30
log 3 = 0.48 [2025]

1 0, 21.84 kJ, –1.26 kJ

2 0, –17.18 kJ, 1.718 J

3 0, 21.84 kJ, 21.84 kJ

4 0, 1.78 kJ, –1.78 kJ

Ans.
(4)

$w = - 2.303 n R T \log_{10} \frac{V_{2}}{V_{1}}$

$w = - 2.303 \times 1 \times 8.3 \times 300 \log_{10} \frac{20}{10} J$

$w = - 2.303 \times 1 \times 8.3 \times 300 \log_{10} 2 J$

$w = - 2.303 \times 1 \times 8.3 \times 300 \times 0.3 J$

$w = - 1720.341 = - 1.72 kJ$

For an ideal gas undergoing any process: $Δ U = n C_{v} Δ T$

For isothermal process $Δ T = 0$ . So $Δ U = 0$ .

By first law of thermodynamics: $Δ U = q + w$

$0 = q + w$

$q = - w = - (- 1.72 kJ) = 1.72 kJ$

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