Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until total volume is 10 litres. How much heat is absorbed and how much work is done in expansion?
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V1 = 2 Litres P1 = 10 atm V2 = 10 litres
T = constant as isothermal expansion.
Since, the temperature is constant, the change in internal energy = 0.
So work done = heat absorbed.
P V = constant = K => P = K / V
K = P1 V1 = 2 litres * 10 atm = 20 atm-litres
![W=\int\limits^{V_2}_{V_1} {P} \, dV\\\\=\int\limits^{V_2}_{V_1} {\frac{K}{V}} \, dV\\\\=K[Ln\ V]_{V_1}^{V_2}\\\\=K*Ln\ \frac{V_2}{V_1}\\\\W=20\ Ln\ \frac{10}{2}=20\ Ln\ 5\ atm-litres](https://tex.z-dn.net/?f=W%3D%5Cint%5Climits%5E%7BV_2%7D_%7BV_1%7D+%7BP%7D+%5C%2C+dV%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7BV_2%7D_%7BV_1%7D+%7B%5Cfrac%7BK%7D%7BV%7D%7D+%5C%2C+dV%5C%5C%5C%5C%3DK%5BLn%5C+V%5D_%7BV_1%7D%5E%7BV_2%7D%5C%5C%5C%5C%3DK%2ALn%5C+%5Cfrac%7BV_2%7D%7BV_1%7D%5C%5C%5C%5CW%3D20%5C+Ln%5C+%5Cfrac%7B10%7D%7B2%7D%3D20%5C+Ln%5C+5%5C+atm-litres "W=\int\limits^{V_2}_{V_1} {P} \, dV\\\\=\int\limits^{V_2}_{V_1} {\frac{K}{V}} \, dV\\\\=K[Ln\ V]_{V_1}^{V_2}\\\\=K*Ln\ \frac{V_2}{V_1}\\\\W=20\ Ln\ \frac{10}{2}=20\ Ln\ 5\ atm-litres")
1 atm = 1.01325 * 10⁵ Pa or N/m²
1 litre = 1000 cc = 10⁻³ m³
Work done = heat absorbed = 20 * 1.01325 * 10⁵ * 10⁻³ Joules
= 2, 026.5 Joules or 2.0265 kilo Joules
T = constant as isothermal expansion.
Since, the temperature is constant, the change in internal energy = 0.
So work done = heat absorbed.
P V = constant = K => P = K / V
K = P1 V1 = 2 litres * 10 atm = 20 atm-litres
1 atm = 1.01325 * 10⁵ Pa or N/m²
1 litre = 1000 cc = 10⁻³ m³
Work done = heat absorbed = 20 * 1.01325 * 10⁵ * 10⁻³ Joules
= 2, 026.5 Joules or 2.0265 kilo Joules
kvnmurty:
forgot to multiply with Ln 5 : Work done= heat absorbed = 3.26 kJoules
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