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An ideal gas undergoes change in its state from the initial state $I$ to the final state $F$ via two possible paths as shown below. Then,
there is no change in internal energy along path $1$
heat is not absorbed by the gas in both paths
the temperature of the gas first increases and then decreases for path $2$
work done by the gas is larger in path $1$
Solution
$(a,c)$ Given paths over $p-V$ graph are
As in both processes, initial and final points are same, change in internal cnergy (which is not a path function) is same. So, option $(a)$ is correct.
– Both processes are expansion process. So, heat is absorbed in both $1$ and $2$ . Hence, option $(b)$ is incorrect.
– Now, consider isotherms over $p-V$ graph given, we clearly see that for path $2$ temperature increases and then decreases. So, option $(c)$ is correct.
– Area enclosed by path $2$ in $p-V$ graph is larger.
So, work done is more in path $2$. Hence, option $(d)$ is incorrect.
