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NEET CHEMISTRYEasy

When 1 mol gas is heated at constant volume, the temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. The correct statement among the following is:

A

q = w = 500 J, \Delta U = 0

B

q = \Delta U = 500 J, w = 0

C

q = 0, w = 500 J, \Delta U = 0

D

\Delta U = 0, q = w = – 500 J

Step-by-Step Solution

  1. Identify Process: The gas is heated at constant volume (isochoric process).
  2. Calculate Work Done (w): Work done in a thermodynamic process is given by w=−PextΔVw = -P_{ext}\Delta V. Since the volume is constant, the change in volume ΔV=0\Delta V = 0. Therefore, work done w=0w = 0.
  3. Apply First Law of Thermodynamics: The law states ΔU=q+w\Delta U = q + w, where ΔU\Delta U is the change in internal energy and qq is the heat supplied.
  4. Substitute Values:
  • Heat supplied (qq) = +500 J+500\text{ J} (positive because heat is absorbed by the system).
  • Work (ww) = 00. ΔU=500 J+0=500 J\Delta U = 500\text{ J} + 0 = 500\text{ J}
  1. Conclusion: q=500 Jq = 500\text{ J}, ΔU=500 J\Delta U = 500\text{ J}, and w=0w = 0.
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