Question 776: Thermodynamics & KTG

Question

For an ideal gas, in an isobaric process, the ratio of heat supplied ' $Q$ ' to the work done ' $w$ ' by the system is ( $\gamma=$ ratio of specific heat at constant pressure to that at constant volume)

Accepted Answer

In an isobaric process (constant pressure), the relationship between the heat supplied ($Q$) and the work done ($w$) by an ideal gas can be analyzed using the first law of thermodynamics and the specific heats.
The first law of thermodynamics states:
$$ Q = \Delta U + w $$
For an ideal gas, the change in internal energy ($\Delta U$) is given by:
$$ \Delta U = nC_v\Delta T $$
where $C_v$ is the specific heat at constant volume.
The work done ($w$) during an isobaric process is:
$$ w = P\Delta V = nR\Delta T $$
where $R$ is the ideal gas constant.
In an isobaric process, the heat supplied ($Q$) can be expressed using the specific heat at constant pressure ($C_p$):
$$ Q = nC_p\Delta T $$
The ratio of specific heats is defined as:
$$ \gamma = \frac{C_p}{C_v} $$
Using the expression for $Q$ and rearranging, we can find the ratio $\frac{Q}{w}$:
$$ \frac{Q}{w} = \frac{nC_p\Delta T}{nR\Delta T} = \frac{C_p}{R} $$
From the definition of $\gamma$, we can write:
$$ C_p = \gamma C_v $$
And from the relationship $C_p - C_v = R$, we have:
$$ C_p = \frac{\gamma R}{\gamma - 1} $$
Thus, the ratio of heat supplied to work done becomes:
$$ \frac{Q}{w} = \frac{C_p}{R} = \frac{\gamma R}{(\gamma-1)R} = \frac{\gamma}{\gamma - 1} $$
Therefore, the correct option is:
Option C
$$ \frac{\gamma}{\gamma-1} $$

Answer

$\frac{1}{\gamma}$

Answer

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

Answer

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

Answer

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

Thermodynamics & KTG

For an ideal gas, in an isobaric process, the ratio of heat supplied ' Q ' to the work done ' w ' by the system is ( \gamma= ratio of specific heat at constant pressure to that at constant volume)

Choose the correct answer: