Question

A gas expands from a volume of \(1\) dm\(^3\) to \(1.25\) dm\(^3\) under a pressure of \(1\) bar. Find the change in internal energy if \(150\) J of heat is supplied to the system.

• A. \(125\) J
• B. \(175\) J
• C. \(-125\) J
• D. \(-175\) J

Hint:

Work done by the gas is \(P\Delta V\); then use the first law \(\Delta U = q + w\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The first law of thermodynamics links the change in internal energy to heat and work: \(\Delta U = q + w\).
Here \(q\) is heat absorbed by the system and \(w\) is work done on the system. When a gas expands against a constant external pressure, the work done on the gas is negative: \(w = -P\Delta V\).

Step 2: Find the Work:
\(\Delta V = 1.25 - 1 = 0.25\) dm\(^3\), and \(P = 1\) bar.
\[ w = -P\Delta V = -(1\ \text{bar})(0.25\ \text{dm}^3) = -0.25\ \text{bar dm}^3 \]
Since \(1\) bar dm\(^3\) = \(100\) J, we get \(w = -25\) J. The sign is negative because the gas does work on the surroundings.

Step 3: Apply the First Law:
Heat is supplied to the system, so \(q = +150\) J.
\[ \Delta U = q + w = 150 + (-25) = 125\ \text{J} \]

Step 4: Why the other options are wrong:
\(175\) J comes from adding \(25\) J instead of subtracting it, which treats the expansion work as work done on the gas. The negative values \(-125\) J and \(-175\) J would mean the internal energy falls, but the system gains \(150\) J of heat and loses only \(25\) J as work, so it must rise.

Final Answer:
The internal energy of the gas rises by \(125\) J, which is option (A). \[ \boxed{125\ \text{J}} \]