Q.255. Thermodynamics - First Law application

Question 255.
Four liters of a monatomic ideal gas, originally at 470 K and a pressure of 3 atm (called state 1), undergoes the following processes:
1 → 2 isothermal expansion to V2 = 7V1
2 → 3 isobaric compression
3 → 1 adiabatic compression to its original state
Find the pressure, volume, and temperature of the gas in state 2 and 3.

Answer 255.
4 liters at 470 K and 3 atm
= 4 * (273/470) * (3/1) liter at NTP
= 6.970 liter
22.4 liter at NTP = 1 mole of gas
=> quantity of gas
= (6.970/22.4) = 0.311 mole.
 Gas in State 2:

V2 = 28 liter
P2 = 3/7 atm
T2 = 470 K

Gas in State 3:

P3 = 3/7 atm
Considering adiabatic compresion 3 to 1,
P1V1^γ = P3V3^γ
=> V3
= V1 * (P1/P3)^(1/γ)
= 4 * (3/(3/7)^(1/1.4)
= 16.06 liter

Also,
P1V1/T1 = P3V3/T3
=> T3
= T1 * ((P3V3) / (P1V1)
= 470 * (3/7) * (16.06) / [3 * 4)
= 269.6 K.

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