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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

A thermodynamic process , the pressures and volumes corresponding to some points in the figure are $\;P_{A} = 3 \times 10^{4}\;pa \;,V_{A} = 2 \times 10^{-3} \;m^3\;$ . $\;P_{B}=8 \times 10^{4}\;pa\;V_{D} = 5 \times 10^{-3}\;m^3\;$. In the process AB , 600 J of heat and in the process BC , 200 J of heat is added to the system . The change in internal energy in process AC would be

$(a)\;560\;J\qquad(b)\;800\;J\qquad(c)\;600\;J\qquad(d)\;640\;J$

1 Answer

Answer : $\;560\;J$
Explanation :
For process AB , W=0 , $\; \bigtriangleup Q =\bigtriangleup u $
Therefore , $\;\bigtriangleup u = 600\;J$
For process BC , $\;\bigtriangleup Q = 200\;J \;,W=P\bigtriangleup V = 8 \times 10^{4} \;[5 \times 10^{-3} - 2 \times 10^{-3}]$
$\bigtriangleup W = 240 J$
$\bigtriangleup Q = \bigtriangleup u + \bigtriangleup W$
$200 = \bigtriangleup u + 240$
$\bigtriangleup u = -40\;J$
Therefore , $\;\bigtriangleup u $\; along AC $\;[\bigtriangleup u \;is\;path\;independent]$
$=600-40$
$ = 560 J$
answered Apr 9, 2014 by yamini.v
 

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