In a given process on an ideal gas , dW=0 and dQ<0. then for the gas
mohit , 11 Years ago
Grade 8
Mechanics> mechanics...
2 Answersneeraj agarwal
Last Activity: 11 Years ago
temperature will decrease as change in dQ=dW+dU dU<0 dU= nfRdT/2 dT<0
ankit singh
Last Activity: 4 Years ago
The work done by the gas or work done on the gas is \((dW = 0)\) zero (which means gas neither expanded nor contracted) but gas lost some \((dQ0)\) energy. This can be compared to air in airtight tube. It's pressure/temperature should decreases but volume remains constant.
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