Question

When an ideal monatomic gas is heated at constant pressure, the fraction of heat energy supplied which increases the internal energy of gas is
 $ \left( A \right)\dfrac{2}{5} \\
  \left( B \right)\dfrac{3}{5} \\
  \left( C \right)\dfrac{3}{7} \\
  \left( D \right)\dfrac{5}{7} \\ $

Answer 1

Hint :Pressure and volume maybe different but temperature should be same at the end state of this process. The internal energy of the ideal gas only depends on temperature. the fraction of heat energy supplied is equal to the ratio of internal energy to the heat supplied. Heat supplies at constant pressure is equal to the work done plus internal energy raised in the process. Also the fraction of heat is equal to the rise in U divided by heat supplied.

Complete Step By Step Answer:
In order to solve this, we first see the first law of thermodynamics;
 $ Q = U + W $
Heat energy supplied at constant volume will be used to raise the internal energy by increasing one degree.
 $ {Q_v} = {C_v}\vartriangle T $
Also when heat is added to the constant pressure, this will be used to increase the internal energy and for the expansion of the work done that is used to maintain the constant pressure which has to be expanded by increasing one degree.
 $ {Q_p} = {C_p}\vartriangle T = W + U $
Pressure and volume may be different but temperature should be the same at the end state of this process. The internal energy of the ideal gas only depends on temperature.
Internal energy is equal to $ \dfrac{f}{2}nRT $ .
So,
 $ Q = \dfrac{f}{2}nRT + P\int {dV} \\
  Q = \dfrac{f}{2}nRT + PV \\ $
Also $ PV $ is equal to $ nRT $ .
Hence;
 $ Q = \dfrac{f}{2}nRT + nRT \\
   = \dfrac{3}{2}nRT + nRT \\
   = \dfrac{5}{2}nRT \\ $
Therefore the fraction of heat energy supplied is equal to the ratio of internal energy to the heat supplied.
 $ = \dfrac{U}{Q} $
Now putting above values we get;
 $ = \dfrac{{(\dfrac{3}{2})nRT}}{{(\dfrac{5}{2})nRT}} \\
   = \dfrac{3}{5} \\ $
So the fraction of the heat energy supplied which increases the internal energy of the gas is $ \dfrac{3}{5} $ .
Hence, option B is the correct option.

Note :
Heat supplies at constant pressure is equal to the work done plus internal energy raised in the process. Also the fraction of heat is equal to the rise in U divided by heat supplied. Pressure and volume may be different but temperature should be the same at the end state of this process. The internal energy of the ideal gas only depends on temperature.