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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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2.5 mol of ideal gas at 2 atm and $27^{\circ}C$ expands isothermally to 2.5 times of its original volume against the external pressure of 1 atm. Calculate q

$\begin {array} {1 1} (A)\;4698.44 \: J & \quad (B)\;-4698.44 \: J \\ (C)\;-5714.15 \: J & \quad (D)\;5714.15 \: J \end {array}$

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1 Answer

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$w = -P(V_2 – V_1)$
Now, initial volume,$(V_1) = \large\frac{nRT+}{P}$$ =\large\frac{ 2.5 \times 0.082 \times 300}{2}$$ = 30.75 \: L$
Final volume, $V_2 = 30.75 \times 2.5 = 76.87 \: L$
Now, $w = -1(76.87 – 30.75) = -46.37 \: L \: atm$
$= -46.37 \times 101.325\: J = -4698.44\: J$
Since the change is isothermal, $ \Delta U = 0$
Now, $ \Delta U = q + w $ or $ 0 = q + w \: or \: q = -w$
$q = -(-4698.44) = 4698.44\: J$
Ans : (A)
answered Mar 16, 2014 by thanvigandhi_1
 

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