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 \((R=8.3\,\text{J mol}^{-1}\text{K}^{-1})\)

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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.
Updated On: Jun 18, 2026
  • \(2500\,\text{J}\)
  • \(1450\,\text{J}\)
  • \(1245\,\text{J}\)
  • \(555\,\text{J}\)
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The Correct Option is C

Solution and Explanation

Step 1: Use the work done formula for a constant pressure process.
For an isobaric process, \[ W=P\Delta V \] Using the ideal gas equation, \[ PV=nRT \] At constant pressure, \[ P\Delta V=nR\Delta T \] Therefore, \[ W=nR\Delta T \]

Step 2: Calculate the change in temperature.

Initial temperature, \[ T_1=30^\circ\text{C} \] Final temperature, \[ T_2=130^\circ\text{C} \] Hence, \[ \Delta T=T_2-T_1 \] \[ \Delta T=130-30 \] \[ \Delta T=100\,\text{K} \]

Step 3: Substitute the given values.

Given, \[ n=1.5 \] \[ R=8.3\,\text{J mol}^{-1}\text{K}^{-1} \] \[ \Delta T=100\,\text{K} \] Therefore, \[ W=nR\Delta T \] \[ W=1.5\times 8.3\times 100 \] \[ W=1245\,\text{J} \]

Step 4: Final conclusion.

Hence, the work done by the gas is \[ \boxed{1245\,\text{J}} \]
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