In an adiabatic process, an ideal gas changes without exchanging heat with the surroundings. The work done in such a process can be understood by looking at the equation for an adiabatic process:
PVγ = constant,
where P is the pressure, V is the volume, and γ (gamma) is the adiabatic index or specific heat ratio (Cp/Cv).
The work done (W) in an adiabatic process can be expressed by the relation:
W = (P1V1 - P2V2) / (γ - 1).
Using the adiabatic condition, one can also express the work done in terms of the change in volume:
The relation simplifies to:
W = (Cv(T1 - T2))/(1-γ)
However, directly relating work done to volume change:
W = ((P1V1 - P2V2))/(γ - 1)) = K(V21-γ - V11-γ)/(1-γ)
Thus, it is clear that the work done depends directly on the volume of the gas during the adiabatic change. Therefore, the correct choice regarding what the work done depends upon in an adiabatic change is: change in its volume.
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Two vessels A and B are connected via stopcock. Vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true?
Choose the correct nuclear process from the below options:
\( [ p : \text{proton}, n : \text{neutron}, e^- : \text{electron}, e^+ : \text{positron}, \nu : \text{neutrino}, \bar{\nu} : \text{antineutrino} ] \)
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: