Question

During a process, system absorbs 710 J of heat and increases the internal energy by 460 J. What is the work performed by system?

• A. $-250\ \text{J}$
• B. $-1170\ \text{J}$
• C. $-710\ \text{J}$
• D. $-460\ \text{J}$

Hint:

Always read the problem text carefully to check the sign conventions! Because the system's internal energy increased ($+460\ \text{J}$) by less than the total heat it absorbed ($+710\ \text{J}$), the missing energy must have been lost as work done by the system, meaning $W$ has to be negative!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to compute the work done during a thermodynamic transformation where the heat energy absorbed and the net change in internal energy are given.

Step 2: Key Formula or Approach:
According to the First Law of Thermodynamics, the change in internal energy ($\Delta U$) is equal to the sum of the heat exchange ($Q$) and work interactions ($W$):
$$\Delta U = Q + W$$ Using standard IUPAC sign conventions in chemical thermodynamics:
Heat absorbed by the system is positive ($Q > 0$).
An increase in internal energy means $\Delta U$ is positive ($\Delta U > 0$).
Work done on the system is positive, and work done by the system is negative.

Step 3: Detailed Explanation:
Let's substitute our given values into the equation:
Heat absorbed ($Q$) = $+710\ \text{J}$
Change in internal energy ($\Delta U$) = $+460\ \text{J}$
Plugging these into our equation:
$$460 = 710 + W$$ Isolate $W$ by moving 710 to the left side:
$$W = 460 - 710$$ $$W = -250\ \text{J}$$ Since the result is negative, it confirms that work is performed

by the system on its surroundings (expansion work). This perfectly matches option (A).

Step 4: Final Answer:
The work performed by the system is $-250\ \text{J}$, which corresponds to option (A).