Question

The work done during adiabatic expansion or compression of an ideal gas is given by:
(the question has multiple correct options)
(A) $n{{C}_{v}}\Delta T$
(B) $\dfrac{nR}{\gamma -1}({{T}_{2}}-{{T}_{1}})$
(C)$-nR{{P}_{ext}}[\dfrac{{{T}_{2}}{{P}_{1}}-{{T}_{1}}{{P}_{2}}}{{{P}_{1}}{{P}_{2}}}]$
(D) $-2.303RT\dfrac{{{V}_{2}}}{{{V}_{1}}}$

Answer 1

In thermodynamics, the subject which is under investigation or collection of matter is being studied is called a system, which is surrounded by a boundary is a closed surface. Through the system and surroundings, the energy and mass changes may be entered or leave.

Complete step by step solution:
According to the thermodynamics in system and surrounding, there are three types of systems,
(1) open system: A system can change both energy and matter with surroundings.
(2) closed system: A system can exchange only energy, not matter with surroundings
(3) isolated system: the system either energy or matter with surroundings.
When an ideal gas compressed adiabatically (Q=0) temperature increases and work is done.
Similarly, adiabatic expansion, the temperature drops with work done.
In the adiabatic process of expansion, with Q=0, and W = 0, according to the first law of thermodynamics, there is no change in the internal energy of the system.
The work is done by the adiabatic expansion, dW=pdV where dQ=0
Therefore, apply the first law of thermodynamics, work done during adiabatic expansion is
$W=n{{C}_{v}}\Delta T$
From the ideal gas law, the temperature of the mixture after the adiabatic compression is
${{T}_{2}}=(\dfrac{{{P}_{2}}{{V}_{2}}}{{{P}_{1}}{{V}_{1}}}){{T}_{1}}$
Then the work done by the mixture during the adiabatic compression is,
$W=\int{{{^{{{V}_{2}}}}_{{{V}_{1}}}}pdV}$
Then W = $-2.303RT\dfrac{{{V}_{2}}}{{{V}_{1}}}$
When an ideal gas in a quasi-static adiabatic process, $P{{V}^{\gamma -1}}=cons\tan t$
A reversible adiabatic expansion of an ideal gas, the work done is,
${{W}_{ext}}=-nR{{P}_{ext}}[\dfrac{{{T}_{2}}{{P}_{1}}-{{T}_{1}}{{P}_{2}}}{{{P}_{1}}{{P}_{2}}}]$

Hence, the options A, C, and D are correct answers.

Note: Most materials expand when heated expect water between ${{0}^{o}}C-{{4}^{o}}C$ , which with increasing temperature decreases volume. ${{C}_{p}}>{{C}_{v}}$ , then expansion occurs under constant pressure with work on surroundings.