ENGR217 2025 resit Q1Past paperOld spec ENGR2172:120 marks24 min

ENGR217 Resit 2025 Q1[VALID] official answers

The cylinder-piston system's walls and piston are made of rigid adiabatic material. Side A and Side B contain N2 and O2, respectively, which have the same temperature, pressure, and volume initially. Given that TA1=TB1=300 KT_{A1} = T_{B1} = 300\ \mathrm{K}, pA1=pB1=0.1 MPap_{A1} = p_{B1} = 0.1\ \mathrm{MPa}, VA1=VB1=0.5 m3V_{A1} = V_{B1} = 0.5\ \mathrm{m^3}. The piston can move freely without friction within the cylinder. After the electric heater on side A is powered on, it slowly heats side A until pA2=0.22 MPap_{A2} = 0.22\ \mathrm{MPa}. Assume that both gases are ideal gases and calculate using constant specific heats: (see Figure Q1-1). (MN2=28.0×10−3 kg/molM_{N_2} = 28.0 \times 10^{-3}\ \mathrm{kg/mol}, MO2=32.0×10−3 kg/molM_{O_2} = 32.0 \times 10^{-3}\ \mathrm{kg/mol}, R=8.3145 J/(mol⋅K)R = 8.3145\ \mathrm{J/(mol\cdot K)}, cV,N2=742.1 J/(kg⋅K)c_{V,N_2} = 742.1\ \mathrm{J/(kg\cdot K)}, cV,O2=649.6 J/(kg⋅K)c_{V,O_2} = 649.6\ \mathrm{J/(kg\cdot K)}, adiabatic index γ=1.4\gamma = 1.4)

Figure Q1-1: Schematic of a piston based cylinder.
Figure Q1-1: Schematic of a piston based cylinder.
Formulas you may need
  • Ideal gas: pV=mRgTpV = mR_gT with Rg=R/MR_g = R/M (and pV=nRTpV = nRT) (on the formula sheet)
  • Adiabatic (reversible) process: pVγ=constpV^\gamma = \mathrm{const}, T2T1=(p2p1)(γ−1)/γ\dfrac{T_2}{T_1} = \left(\dfrac{p_2}{p_1}\right)^{(\gamma-1)/\gamma} (on the formula sheet)
  • First law: dU=dQ−dWdU = dQ - dW (heat in +, work out +), dU=mcv dTdU = mc_v\,dT (on the formula sheet)
  • Adiabatic work: W=−ΔUW = -\Delta U; polytropic work magnitude p2V2−p1V1n−1\dfrac{p_2V_2 - p_1V_1}{n - 1} with n=γn = \gamma (on the formula sheet)
  • Free frictionless piston: equal pressures on both sides at every instant (learn this)
  1. (a)
    TB2T_{B2} and VB2V_{B2}
    [7]
  2. (b)
    VA2V_{A2} and TA2T_{A2}
    [7]
  3. (c)
    QQ and WAW_A (work done by side A on side B)
    [6]