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A gas is heated at a constant pressure. The fraction of heat supplied used for external work is:
(A) \[\dfrac{1}{\gamma }\]
(B) \[1 - \dfrac{1}{\gamma }\]
(C) \[\gamma - 1\]
(D) \[1 - \dfrac{1}{{{\gamma ^2}}}\]

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Last updated date: 17th Apr 2024
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Answer
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Hint: Such a system must obey the first law of thermodynamics. Use the equation of the first law of thermodynamics to find the work done by the system.
Formula used: In this solution we will be using the following formulae;
\[\Delta U = \Delta Q - W\] where \[\Delta U\] is the change in internal energy of the system, \[\Delta Q\] is the change in thermal energy of (or the heat absorbed by) the system, and \[W\] is the work done by the system.
\[\Delta Q = m{c_p}\Delta T\] where \[m\]is the mass of the gas, \[{c_p}\]is the specific heat at constant pressure, and \[\Delta T\] is the change in temperature of the system.
\[\Delta U = m{c_v}\Delta T\] where \[{c_v}\] is the specific capacity at constant volume.

Complete Step-by-Step Solution:
When the gas is heated, the change in thermal energy would be given by
\[\Delta Q = m{c_p}\Delta T\]where \[m\] is the mass of the gas, \[{c_p}\] is the specific heat at constant pressure, and \[\Delta T\] is the change in temperature of the system.
At the same time, the change in internal energy is given by
\[\Delta U = m{c_v}\Delta T\] where \[{c_v}\] is the specific capacity at constant volume.
Now, from the first law of thermodynamics given as
\[\Delta U = \Delta Q - W\] where \[W\] is the work done by the system, we can find the work done as
\[W = \Delta Q - \Delta U\]
\[\dfrac{W}{{\Delta Q}} = \dfrac{{\Delta Q - \Delta U}}{{\Delta Q}} = 1 - \dfrac{{\Delta U}}{{\Delta Q}}\]
Replacing the known expressions into above equation, we have
\[\dfrac{W}{{\Delta Q}} = 1 - \dfrac{{m{c_v}\Delta T}}{{m{c_p}\Delta T}} = 1 - \dfrac{{{c_v}}}{{{c_p}}}\]
The ratio \[\dfrac{{{c_p}}}{{{c_v}}}\] is usually given the constant \[\gamma \]
Hence,
\[\dfrac{W}{{\Delta Q}} = 1 - \dfrac{1}{\gamma }\]

Thus, the correct option is B

Note: To avoid confusions, the thermodynamic equation can be written as
\[\Delta U = \Delta Q + W\]
However, in this format, the definition of \[W\] is the work done on (not by) the system. Hence, it is negative in value when work is done by the system