ThermodynamicsHard
Question
The work done on one mole of an ideal gas in the reversible process: PV3 = constant, from (1 atm, 300 K) to 2 2 atm is
Options
A.150R
B.300R
C.75R
D.600R
Solution
$P_{1} = 1\text{ atm,}V_{1} = V\text{ L, }\text{T}_{1} = 300\text{ K}$
$P_{2} = 2\sqrt{2}\text{ atm, }V_{2} = V_{1}.\left( \frac{P_{1}}{P_{2}} \right)^{1/3} = V_{1}\left( \frac{1}{2^{3/2}} \right)^{1/3} = \frac{V_{1}}{\sqrt{2}};T_{2} = \frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$
$\text{Or }T_{2} = \frac{2\sqrt{2}}{1} \times \frac{1}{\sqrt{2}} \times 300 = 600\text{ K}$
$\text{Now, }w = \frac{- nR\left( T_{2} - T_{1} \right)}{1 - x} = - \frac{1 \times R \times (600 - 300)}{1 - 3} = + 150\text{ R}$
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