An ideal gas has undergone through the cyclic process as shown in the figure. Work done by the gas in the entire cycle is _____ $ \times 10^{-1} $ J. (Take $ \pi = 3.14 $)
The graph is a circle in the PV diagram, and for a cyclic process, the work done by the gas is equal to the area enclosed by the cycle.
Assuming both axes are scaled equally, the radius \( R = 100 \) \[ \text{Area} = \pi R^2 = 3.14 \times 100^2 = 3.14 \times 10^4 \] Convert units: \[ 1\, \text{kPa} \cdot \text{cm}^3 = 10^{-2} \, \text{J} \Rightarrow \text{Work done} = 3.14 \times 10^4 \times 10^{-2} = 314 \, \text{J} \]
Work done = $31.4 \times 10^{-1}$ J
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: