Two litres of an ideal gas at a pressure of $10 a t m$ expands isothermally into a vacuum until its total volume is $10$ litres. How much heat is absorbed and how much work is done in the expansion?

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Solution

Isothermal Expansion $:\to$
Temperature = constant
$P v = n R T$ (.. universal ideal gas equation)
Let n = constant = 1
R = universal gas constant = 8.314 J/mol.k
T = constant (Isothormal)
So Pv = constant
$T = \frac{P V}{R} = \frac{20}{8.314} \times 101.3$ (in SI)
$P_{1} V_{1} = P_{2} V_{2}$
Let $P_{1} = 10 a t m$ and $P_{2} = ?$
$V_{1} = 2$ lts $1 / 2 = 10$ lts
As $P_{2}$ = final pressure So, $P_{1} V_{1} = P_{2} v_{2}$
$\Rightarrow 10 \times 2 = P_{2} \times 10$
$\Rightarrow P_{2} = 2 a t m$
Acc. to first law of thermodynamics, the expansion
at constant temperature, leads to internal energy (1)
is equal to zero
$Q = A V + N \to (F . L . O . T)$
$Q = N = R T l n \frac{V_{2}}{V_{1}} = 101.3 \times 8.314 \times \frac{20}{8.314} \times l n \frac{10}{2}$
$= 3260.72 J$
Heat absorbed $= + 3260.72 J$
work done by system $= - 3260.72 J$

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Two litres of an ideal gas at a pressure of 10atm expands isothermally into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?

Two litres of an ideal gas at a pressure of $10 a t m$ expands isothermally into a vacuum until its total volume is $10$ litres. How much heat is absorbed and how much work is done in the expansion?