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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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1 mole of ideal monatomic gas at $ 27^{\circ}C$ expands adiabatically against a constant external pressure of $1.5\: atm$ from a volume of $4\: dm^3\: to\: 16\: dm^3$ . Calculate $ \Delta E$.

$\begin {array} {1 1} (A)\;1823.85\: J & \quad (B)\;- 1823.85 \: J \\ (C)\;3039.75 \: J & \quad (D)\;-3039.75 \: J \end {array}$

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Since the process is adiabatic, so, $q=0$
As the gas expands against the constant external pressure
$W = -P (\Delta V) = -1.5 \times (V_2 – V_1) = -1.5(16-4)$
$= -18\: atm \; dm^3 = -18 \times 101.325 \: J = -1823.85 /j$
So, $ \Delta E = q + W = 0 + (-1823.85) = -1823.85\: J$
Ans : (B)
answered Mar 16, 2014 by thanvigandhi_1
 

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