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

When helium gas is heated at constant pressure fraction of heat that goes into increasing the internal energy is ---

$(a)\;\large\frac{2}{5}\qquad(b)\;\large\frac{3}{5}\qquad(c)\;0.5\qquad(d)\;cant\;be\;determined$

1 Answer

Answer : (b) $\;\large\frac{3}{5}$
Explanation :
Total heat $\; \bigtriangleup Q = \bigtriangleup C_{p} \bigtriangleup T$
Internal energy change $\; \bigtriangleup u = n C_{v} \bigtriangleup T$
Therefore , Fraction of energy going to increase internal energy ---
$\large\frac{\bigtriangleup u} { \bigtriangleup Q} = \large\frac{C_{v}}{ \bigtriangleup C_{p} } = \large\frac{1}{r} =\large\frac{3}{5}$
(Since helium is monoatomic )
answered Mar 13, 2014 by yamini.v
 

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