Question 788: Thermodynamics & KTG

Question

In an isobaric process of an ideal gas, the ratio of work done by the system to the heat supplied $\left(\frac{W}{Q}\right)$ is

Accepted Answer

For an isobaric process involving an ideal gas, we need to find the ratio of the work done by the system to the heat supplied, denoted as $\frac{W}{Q}$.
In an isobaric process:
The heat supplied to the system is given by:
$    \Delta Q = n C_p \Delta T    $
where $n$ is the number of moles, $C_p$ is the specific heat at constant pressure, and $\Delta T$ is the change in temperature.
The change in internal energy is:
$    \Delta U = n C_v \Delta T    $
where $C_v$ is the specific heat at constant volume.
The work done by the system is:
$    W = \Delta Q - \Delta U = n C_p \Delta T - n C_v \Delta T    $
To find the ratio $\frac{W}{\Delta Q}$:
$ \begin{aligned} \frac{W}{\Delta Q} &= \frac{n C_p \Delta T - n C_v \Delta T}{n C_p \Delta T} \ &= \frac{C_p - C_v}{C_p} \ &= \frac{\frac{C_p}{C_v} - 1}{\frac{C_p}{C_v}} \end{aligned} $
Since the ratio $\frac{C_p}{C_v} = \gamma$, where $\gamma$ is the heat capacity ratio, we have:
$ \frac{W}{\Delta Q} = \frac{\gamma - 1}{\gamma} $
This concludes the derivation, showing that the ratio of work done by the system to the heat supplied in an isobaric process is $\frac{\gamma - 1}{\gamma}$.

Answer

$\frac{1}{\gamma-1}$

Answer

$\gamma$

Answer

$\frac{\gamma}{\gamma-1}$

Answer

$\frac{\gamma-1}{\gamma}$

Thermodynamics & KTG

In an isobaric process of an ideal gas, the ratio of work done by the system to the heat supplied \left(\frac{W}{Q}\right) is

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