Question:
An ideal monatomic gas expands at constant pressure. The work done by the gas on its environment is \(200\,\text{J}\), then the heat absorbed by the gas during the process is
An ideal monatomic gas expands at constant pressure. The work done by the gas on its environment is \(200\,\text{J}\), then the heat absorbed by the gas during the process is
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For a monatomic ideal gas, \(\Delta U=\frac{3}{2}nR\Delta T\). At constant pressure, \(W=nR\Delta T\), so \(Q=\Delta U+W\).
Updated On: Jun 26, 2026
- \(500\,\text{J}\)
- \(300\,\text{J}\)
- \(200\,\text{J}\)
- \(600\,\text{J}\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the relation for work done at constant pressure.
For an ideal gas expanding at constant pressure, \[ W=P\Delta V \] Given, \[ W=200\,\text{J} \] For an ideal gas, \[ P\Delta V=nR\Delta T \] Hence, \[ nR\Delta T=200\,\text{J} \]
Step 2: Find change in internal energy.
For a monatomic ideal gas, \[ \Delta U=\frac{3}{2}nR\Delta T \] Since \[ nR\Delta T=200, \] we get \[ \Delta U=\frac{3}{2}\times 200 \] \[ \Delta U=300\,\text{J} \]
Step 3: Apply the first law of thermodynamics.
According to the first law, \[ Q=\Delta U+W \] So, \[ Q=300+200 \] \[ Q=500\,\text{J} \]
Step 4: Final conclusion.
Hence, the heat absorbed by the gas is \[ \boxed{500\,\text{J}} \]
For an ideal gas expanding at constant pressure, \[ W=P\Delta V \] Given, \[ W=200\,\text{J} \] For an ideal gas, \[ P\Delta V=nR\Delta T \] Hence, \[ nR\Delta T=200\,\text{J} \]
Step 2: Find change in internal energy.
For a monatomic ideal gas, \[ \Delta U=\frac{3}{2}nR\Delta T \] Since \[ nR\Delta T=200, \] we get \[ \Delta U=\frac{3}{2}\times 200 \] \[ \Delta U=300\,\text{J} \]
Step 3: Apply the first law of thermodynamics.
According to the first law, \[ Q=\Delta U+W \] So, \[ Q=300+200 \] \[ Q=500\,\text{J} \]
Step 4: Final conclusion.
Hence, the heat absorbed by the gas is \[ \boxed{500\,\text{J}} \]
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