N moles of CO2 gas (T1, V1) obey a van der Waals equation of state are contained in an insulated piston-cylinder arrangement. A reversible expansion of the gas is carried out until the volume become 2V1. [CV. m = 28.80 J mol−1K−1, a = 3.610 atm. dm6. mol−2, b = 4.29 x 10−2 dm3. mol−1]
(a) Find the final temperature of gas as a function of n, . , T1, V1 and the van
der Waals parameter a and b.
(b) If two moles of CO2 gas involved and T1 = 350K, V1 = 40L, compute final
temperature of the gas.
(c) Find work done and internal energy of the gas.

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We have no idea what the graph shows, or why it may be accurate,  because we can't see the graph.on a different subject, choice ' d ' is the only true statement among the choices.

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: where m is the mass of the arrow and v its velocity.  all the other options, instead, do not increase the kinetic energy of the system: -  pulling farther back on the string : this will just increase the elastic potential energy of the system-  holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)-  decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)