Question

An ideal monoatomic gas of \(1.5\) moles is heated at a constant pressure of \(2\,\text{atm}\) so that its temperature increases from \(30^\circ\text{C}\) to \(130^\circ\text{C}\). Work done by the gas is \((R=8.3\,\text{J mol}^{-1}\text{K}^{-1})\)

• A. \(2500\,\text{J}\)
• B. \(1450\,\text{J}\)
• C. \(1245\,\text{J}\)
• D. \(555\,\text{J}\)

Hint:

For an ideal gas undergoing an isobaric process, \[ W=P\Delta V=nR\Delta T. \] You do not need the pressure value explicitly once \(n\), \(R\), and \(\Delta T\) are known.

Answer 1

The Correct Option is C

Step 1: Use isobaric work formula.
At constant pressure, W = PΔV. From ideal gas law, PΔV = nRΔT.

Step 2: Compute temperature change.
ΔT = 130 – 30 = 100 K.

Step 3: Calculate work done.
W = nRΔT = 1.5 × 8.3 × 100 = 1245 J.

Step 4: Final Answer:
1245 J.