To solve the given problem, we need to analyze the changes in the thermodynamic parameters of the system when a liquid in a thermally insulated closed vessel is mechanically stirred. The key steps involve understanding the implications of each listed parameter under the given conditions:
1. System Description: The liquid is in a thermally insulated closed vessel, which implies that there is no heat exchange with the surroundings, i.e., q = 0.
2. Work Done: Mechanical stirring involves external work being done on the system. Thus, w > 0, as work is added to the system from the surroundings.
3. Change in Internal Energy (∆U): According to the first law of thermodynamics: \(\Delta U = q + w\). Substituting the known values, we get:
\(\Delta U = 0 + w \gt 0\)
Since w > 0, the internal energy of the system increases, making ∆U > 0.
Therefore, the correct option for the thermodynamic parameters is:
\(\Delta U>0, q=0, w>0\)
The left and right compartments of a thermally isolated container of length $L$ are separated by a thermally conducting, movable piston of area $A$. The left and right compartments are filled with $\frac{3}{2}$ and 1 moles of an ideal gas, respectively. In the left compartment the piston is attached by a spring with spring constant $k$ and natural length $\frac{2L}{5}$. In thermodynamic equilibrium, the piston is at a distance $\frac{L}{2}$ from the left and right edges of the container as shown in the figure. Under the above conditions, if the pressure in the right compartment is $P = \frac{kL}{A} \alpha$, then the value of $\alpha$ is ____
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
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: