Question 781: Thermodynamics & KTG

Question

The change in the internal energy of the mass of gas, when the volume changes from ' $V$ ' to ' 2 V ' at constant pressure ' $P$ ' is ( $\gamma$ is the ratio of specific heat of gas at constant pressure to specific heat at constant volume)

Accepted Answer

The change in the internal energy, ÃÂU, of a gas when its volume changes from 'V' to '2V' at constant pressure 'P' can be determined using the following derivation.
The formula for the change in internal energy is given by:
$ \Delta U = n C_V \Delta T \quad 	ext{.... (i)} $
We are given that the ratio of specific heats is:
$ \frac{C_P}{C_V} = \gamma $
and the relation between them is:
$ C_P - C_V = R $
This leads to:
$ 1 + \frac{R}{C_V} = \gamma $
From this, we derive:
$ C_V = \frac{R}{\gamma - 1} $
Substitute $ C_V $ into equation (i):
$ \Delta U = n \left(\frac{R}{\gamma - 1}ight) \Delta T $
Under constant pressure, we use the equation:
$ P \Delta V = nR \Delta T $
Thus, substituting for $\Delta T$ from above:
$ \Delta U = \frac{P \Delta V}{\gamma - 1} = \frac{P(2V - V)}{\gamma - 1} = \frac{PV}{\gamma - 1} $
This indicates that the change in internal energy is $ \frac{PV}{\gamma - 1} $.

Answer

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

Answer

$\frac{\mathrm{PV}}{\gamma+1}$

Answer

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

Answer

$\frac{\gamma+1}{\mathrm{PV}}$

Thermodynamics & KTG

The change in the internal energy of the mass of gas, when the volume changes from ' V ' to ' 2 V ' at constant pressure ' P ' is ( \gamma is the ratio of specific heat of gas at constant pressure to specific heat at constant volume)

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