Question

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,

• A. there is no change in internal energy along path $1$
• B. heat is not absorbed by the gas in both paths
• C. the temperature of the gas first increases and then decreases for path $2$
• D. work done by the gas is larger in $1$

Answer 1

Answer: (B), (C)

As in both processes, initial and final points are same, change in internal energy (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.