The work done (in Cal) in adiabatic compression of 2 moles of an ideal monatomic gas by the constant external pressure of 2 atm starting from an initial pressure of 1 atm and an initial temperature of

Question

The work done (in Cal) in adiabatic compression of 2 moles of an ideal monatomic gas by the constant external pressure of 2 atm starting from an initial pressure of 1 atm and an initial temperature of 300 K is: [ R = 2 cal / mol − K ]

TestHub · Accepted Answer

To calculate the work done, we neet to know the final temperature.

$T_{1} = 300 K, P_{1} = 1 atm, n = 2 mole, P_{2} = 2 atm, V_{1} = \frac{{nRT}_{1}}{P_{1}}$

$w = Δ Y = {nC}_{v} (T_{2} - T_{1}) = - P_{2} (V_{2} - V_{1})$

For a monatomic gas, $C_{v} = \frac{3}{2} R$

$\Rightarrow \frac{3}{2} nR (T_{2} - T_{1}) = - P_{2} (V_{2} - V_{1})$

$\frac{3}{2} nR (T_{2} - T_{1}) = - P_{2} (nR \frac{T_{2}}{P_{2}} - nR \frac{T_{1}}{P_{1}})$

$\frac{3}{2} (T_{2} - T_{1}) = - T_{2} + \frac{P_{2}}{P_{1}} \Rightarrow T_{2} = \frac{2}{3} (\frac{P_{2}}{P_{1}} + \frac{3}{2}) T_{1}$

$T_{2} = 0 .4 (2 + 1 .5) 300 = 420 K$

Work done, $W = {nC}_{v} ΔΤ = 2 \times \frac{3}{2} R \times (420 - 300)$

$\Rightarrow W = 2 \times \frac{3}{2} \times 2 \times 120 = 720 cal$

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