Ten moles of an ideal monoatomic gas, initially in state a at atmospheric pressure and temperature Ta=27 , is enclosed in a metal cylinder of volume V0 fitted with a frictionless piston. The gas is suddenly compressed to state b with volume V0/3. Now, keeping the piston stationary, the cylinder is submerged in a water bath of temperature 11 until the gas reaches the temperature of the water bath, which is denoted as state c. Finally, while still in the water bath, the piston is brought slowly to its initial position, which is denoted as state f. If R is universal gas constant, then the correct option(s) is/are: [Given: 91/3=2.08]
- The schematic P-V diagram of the processes described above is: [figure]
- The change in internal energy in going from state a to b is 4860R.
- The net change in the internal energy in the whole process is -240R.
- The pressure and temperature of the state b are 2.08 times the atmospheric pressure and 624 K, respectively.
Answer
(1) The schematic P-V diagram of the processes described above is: [figure], (2) The change in internal energy in going from state a to b is 4860R., (3) The net change in the internal energy in the whole process is -240R.
Solution
Key idea: a b adiabatic ( = 53), b c isochoric cooling, c f isothermal expansion at 284 K.
Tb=Ta ( V0V0/3 ) -1=300 32/3=300 91/3=624 K; Pb=Pa 35/3=3 2.08 Pa=6.24 Pa. (D) false.
Uab=nCV T=10 32R 324=4860R. (B) true.
Net: U=10 32R (284-300)=-240R. (C) true.
(A): adiabat from a (at V0) to b (at V0/3, higher P), vertical drop b c, then isotherm c f back to V0 ending below a since Tf<Ta — matches the diagram. True.
Answer: (A), (B), (C)