A system comprises 7.14 grams of Ne gas at 0 degree C and 1 atm. When 2025 joule

Question

A system comprises 7.14 grams of Ne gas at 0 degree C and 1 atm. When 2025 joules of heat are added to the system at constant pressure, the resultant expansion causes the system to perform 810 joules of work. Calculate (a) the initial state, (b) the final state, (c) delta U and delta H for the process, and (d)c_p and c_v. The molecular weight of Ne is 20 and it can be assumed that Ne behaves as an ideal gas. A system comprises one mole of an ideal gas at 0 degree C and 1 atm. The system is subjected to the following processes, each of which is conducted reversibly: A 10-fold increase in volume at constant temperature hen a 100-fold adiabatic increase in pressure then a return to the initial state along a straight-line path in the P-V diagram Calculate the work done by the system in step (c) and the total heat added to or withdrawn from the system as a result of the cyclic process. One mole of an ideal gas at 25 degree C and 1 atm undergoes the following reversibly conducted process: isothermal expansion to 0.5 atm, followed by Isobaric expansion to 100 degree C, followed by Isothermal compression to 1 atm, followed by Isobaric compression to 25 degree C The system then undergoes the following cyclic process: isobaric expansion to 100 degree C, followed by A decrease in pressure at constant volume to P atm, followed by An isobaric compression at P atm to 24.5 liters, followed by An increase in pressure at constant volume to 1 atm Calculate the value of P which makes the work done on the gas in the first cycle equal to the work done by the gas in the second cycle.

Explanation / Answer

For Neon, Molar heat capacities are :

CP=5/2R and CV= 3/2 R

for constant pressure process, deltaU= Q-W ( - sign is there because work is done by the system)

deltaU= 2025-810=1215 joules

Q= deltaH = 810 joules

molar mass of gas = 20, moles of neon = mass/molar mass = 7.14/20 =0.357

volume at initial condition = nRT/P= 0.357*0.0821*273/1 =8 L

for isobaric process, W= -R*(T2-T1)

=-8.314*(T2-273)= 810

273-T2= 810/8.314=97.43

T2= 273-97.42=175.57 K

at final state, since P is constant V1/T1= V2/T2

8/273= V2/175.57

V2 =8*175.57/273=5.14 L

2.

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