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

In a thermodynamic process 400 J of heat is given to a gas and 100 J of work is also done on set . Calculate the change in internal energy of the gas .

$(a)\;800\;J\qquad(b)\;500\;J\qquad(c)\;700\;J\qquad(d)\;600\;J$

1 Answer

Answer : $\;500\;J$
Explanation :
$dQ=du+dw \quad $ By sign convention
$dQ=+400\;J quad$ Heat absorbed is positive
$dw=-100\;J \quad$ heat given out is negative
$du=dQ-dw \quad$ Work done by the system is positive
$=400 - (-100)\quad$ Work done on the system is negative
$=500 \;J$
answered Mar 21, 2014 by yamini.v
 

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