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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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A sample of gas is compressed by an average pressure of $0.5\: atm$ so as to decrease its volume from $400 \: cm^3$ to $200\: cm^3$ . During the process $8\: J$ of heat flows out to surroundings. Calculate the change in internal energy of the system

$\begin {array} {1 1} (A)\;-8 \: J & \quad (B)\;-2.13\: J \\ (C)\;2.13 \: J & \quad (D)\;18.13 \: J \end {array}$

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Here, $ \Delta V = 200-400 = -200\: cm^3 = -0.2L$
External pressure(P) $= 0.5\: atm$
$W = -P \Delta V = -0.5 \times (-0.2) = 0.1\: L-atm$
Now, $1L-atm = 101.3J$
$W = 0.1 \times101.3 = 10.13\: J$
$Q = -8\: J $
$ \Delta U = Q + W= -8 + 10.13\: J = 2.13\: J$
Ans : (C)
answered Mar 15, 2014 by thanvigandhi_1
edited Mar 21, 2014 by balaji.thirumalai
 

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