Question

PLEASE SOLVE ALL QUESTIONS ( EMPHASIZE ON D + E ) PLEASE ( CLEAR HANDWRITING)

Question 2 (each part is worth 5 marks) Data for oxygen gas (02): relative molecular mass: 32.0 molar specific heat at constant pressure: 29.1 J nol-1K-1 molar specific heat at constant volume: 20.8 J mol-1K-1 Assume all gases are ideal (a) 100 mol of a gas (not oxygen) is initially at temperature -20°C. The gas undergoes isobaric expansion, such that the final temperature is +300C. The energy transfer by heating (for the gas) is Q-1.68 x 105 J. Find the molar specific heat at constant pressure. (b) Find the number of moles in 0.1 kg of oxygen gas, giving your answer to 3 significant figures. (c) 100 mol of oxygen gas undergocs isochoric heating. The initial pressure and temperature are P105 Pa, To - 200 K and the final pressure is P - 1.5 x 105 Pa. Find the energy transfer by heating for the gas, giving your answer in scientific (power of ten) notation. (d) When oxygen gas undergoes an adiabatic process, the pressure P and volume V obey the following formula: PV14 constant If the initial pressure and volume are Fo 105 Pa, o0.1 m and the final pressure is B-5x 104 Pa; find the final volume V . (e) Oxygen gas undergoes isobaric expansion due to heating. The energy transfer by heating, for the gas, is Q+4 x 105 J. Find the energy transfer by working, for the gas (Hints: from initial volume to final volume jat constant P. ΡΔν-PV1 ~ Ph. Also, you do not need to know n or P)

Answer 1

(a) We know that
Q = nC_P*(T_final - T_initial)
where n is no. of moles = 100
T_final is final temperature = 30 ^oC = 30+273 = 303 K
T_initial is initial temperature = -20 ^oC = -20+273 = 253 K
Q is heat transferred = 1.68*10⁵ J
1.68*10⁵ = 100*C_P*(303 - 253)
C_P = 33.6 J/mol-K
(b)
We know that
1 mole = 32 gram = 0.032 kg
1 kg = (1/0.032) mole
Now no. of moles in 0.1 kg
0.1 kg = 0.1*(1/0.032) = 3.125 moles
(c)
We know that the heat transferred at constant volume is given by
P/T = Constant
T₁ = (P₁/P_o)*T_o
T₁ = (1.5)*200= 300 K
Q = nC_V*(T_f - T_i)
Q = 100*20.8*(300-200) = 2.08*10⁵ J
(d)
We know that in the adiabatic process
$\frac{V_{1}}{V_{o}}= \left ( \frac{P_{o}}{P_{1}} \right )^{\frac{1}{\gamma }}$
Puttting all the values , we get
$\frac{V_{1}}{0.1}= \left ( \frac{10^{5}}{5*10^{5}} \right )^{\frac{1}{1.4 }}$
On solving , we get
V₁ = 0.0317 m³