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1st Law of Thermodynamics

Two moles of ...

Question

Two moles of an ideal gas $(C_{v} = \frac{5}{2} R)$ was compressed adiabatically against constant pressure of $2 a t m$ , which was initially at $350 K$ and $1 a t m$ pressure. The work involve in the process is equal to?

A

250 R

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B

300 R

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C

400 R

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D

500 R

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Solution

The correct option is D $500 R$
For reversible adiabatic process,

$W = - P_{e x t} [\frac{n R T_{2}}{P_{2}} - \frac{n R T_{1}}{P_{1}}]$
$P_{2} = P_{e x t} = 2 a t m$

$P_{1} = 1 a t m$ $T_{1} = 300 K$

$W = - (2 a t m) [\frac{2 (R) . T_{2}}{2 a t m} - \frac{2 R (350)}{1 a t m}]$

and $W = 2 C_{V} (T_{2} - 350) = 2 \times \frac{5}{2} R (T_{2} - 350)$

$5 R (T_{2} - 350) = (750 R - 2 R T_{2})$

$5 T_{2} - 1750 = 1400 - 2 T_{2}$

$7 T_{2} = 3150$ $T_{2} = 450 K$

$W = 2 \times C_{V} (450 - 350)$

$= 2 \times \frac{5}{2} R \times (100) = 500 R$

$W = W_{A B} + W_{B C} + W_{C D}$
$= - P_{0} (V_{B} - V_{A}) = - n R T_{B} l n [\frac{V_{C}}{V_{B}}] - \frac{P_{0}}{2} (V_{D} - V_{C})$

$W = - P_{0} (2 V_{0} - V_{0}) - 2 P_{0} V_{0}$ ln

$[\frac{4 V_{0}}{2 V_{0}}] - \frac{P_{0}}{2} (2 V_{0} - 4 V_{0})$

$W = - 2 P_{0} V_{0} l n 2$ and $q = - W (∵ Δ U = 0)$

$q = 2 P_{0} V_{0} l n 2$

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