Agas is confined to a container with a massless piston at the top. a massless wire is attached to the piston. when an external pressure of 2.03 bar is applied to the wire, the gas compresses from 6.40 to 3.20 l . when the external pressure is increased to 2.53 bar, the gas further compresses from 3.20 to 2.56 l .

in a separate experiment with the same initial conditions, a pressure of 2.53 bar was applied to the gas, decreasing its volume from 6.40 to 2.56 l in one step.

if the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?

q in case of two -step process will be less in magnitude than q in case of one step process.

Explanation:

Heat q given to a system is used in 1) increasing the internal energy of the gas

and 2) work done in increasing the volume . If there is decrease in volume , work done is negative.

In both the cases change in internal energy of the gas is same as the rise in temperature is same.  But work done which is negative in nature will have greater value in one step process. It is so because work done is equal to external pressure multiplied by change in volume. Change in volume is same but external pressure is more in case of two step process. Hence negative work done is more in case of two step process. Hence q in case of two step process will be lesser.

 q = change in internal energy - work done on the gas.

a, c, d, e are all applying about success strategies.

attach the statements so i can answer : )