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प्रश्न
0.022 kg of CO2 is compressed isothermally and reversibly at 298 K from an initial pressure of 100 kPa when the work obtained is 1200 J, calculate the final pressure.
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उत्तर
Given: Mass of CO2 = 0.022 kg = 22 g
Temperature = T = 298 K
Initial pressure = P1 = 100 kPa
Work done on system = Wmax = 1200 J
To find: Final pressure = P2
Formula: Wmax = – 2.303 nRT log10 `"P"_1/"P"_2`
Calculation: Number of moles of CO2 = n = `(22 "g")/(44 "g mol"^-1)` = 0.5 mol
Gas constant = R = 8.314 J K–1 mol–1
Now, using formula,
Wmax = – 2.303 nRT log10 `"P"_1/"P"_2`
1200 J = − 2.303 × 0.5 mol × 8.314 J K–1 mol–1 × 298 K × log10 `(100 "kPa")/"P"_2`
∴ `log_10 (100 "kPa")/"P"_2 = (- 1200)/(2.303 xx 0.5 xx 8.314 xx 298)`
Calculation using log table:
`(1200)/(2.303 xx 0.5 xx 8.314 xx 298)`
= Antilog10 [log10 1200 – (log10 2.303 + log10 0.5 + log10 8.314 + log10 298)]
= Antilog10 [3.0792 – (0.3623 + `bar"1"`.6990 + 0.9198 + 2.4742)]
= Antilog10 [3.0792 – (3.7563 + `bar"1"`.6990)]
= Antilog10 [`bar"1"`.6239] = 0.4207
= - 0.4207
∴ `(100 "kPa")/"P"_2` = antilog10 (– 0.4207) = 0.3796
∴ `"P"_2 = (100 "kPa")/0.3796`
= 263.4 kPa
The final pressure is 263.4 kPa.
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