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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) $1 - \dfrac{1}{{{\gamma ^2}}}$

Last updated date: 09th Aug 2024
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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