One mole of an ideal gas in a closed system undergoes a mechanically reversible adiabatic compression process and changes from V1= 0.05 m^3 and P1= 1 bar to P2= 12 bar. Calculate Q, W, ∆U, and ∆H of the process. If the process will become irreversible with 50% efficiency, calculate the W, Q, ∆U, and ∆H.
One mole of an ideal gas in a closed system undergoes a mechanically reversible adiabatic compression...
An ideal gas undergoes a cycle consisting of the following mechanically reversible steps: An adiabatic compression from Pu V1, T1 to P2, V2, T2 An isobaric expansion from P2, V2, T2 to P3 P2, Vs, T3 - An adiabatic expansion from P3, Vs, Ts to Pa, V4, T4 - A constant-volume process from Pa, V4, T4 to Pi, VV4, T1 (a) Sketch this cycle on a PV diagram (b) Derive an equation that expresses the thermal efficiency (n) of this...
One mole of an ideal gas undergoes a reversible adiabatic expansion from T_1, to T_2 while tripling the volume of the gas. What is the relation between T_1 and T-2? T-2/3 < T_1<T_2 T_2/3 < T_1 < T-2 T_1= T_2 T_2<T_1 T_1 lessthanorequalto T_2/3 One mole of Ar gas undergoes the reversible transformation shown. Assuming Ar behaves ideally, which statement is true for step 2? Delta U= C_p DeltaT DeltaH < Delta U Delat S= c_p ln(T_c/T_B) W = etaRt...
An ideal gas initially at 600K and 10 bar undergoes a four-step mechanically reversible cycle in a closed system. In step 12 (the process that changes the system from State 1 to State 2), pressure decreases isothermally to 3bar; in step 23, pressure decreases at constant volume to 2bar; in step 34 volume decreases at constant pressure; and in step 41, the gas returns adiabatically to its initial state. Take Cp=(7/2)R and Cv=(5/2)R. Determine the efficiency of the cycle. (Hint:...
12. 1 mole of an ideal gas undergoes an isothermal expansion from V1 = 1.4L followed by isobaric compression, p = cst.if P1 = 4.4atm, p2 = 1.7atm → ?- m calculate the work done by gas during the expansion. Express work in J = N·m! • For isothermal processes, AT = 0 T = cst → w=faw=fr&v=/MRT AV 594 Show your work like: `x-int_0^5 v(t)dt rarr x-int_0^5(-4*t)dt=-50 m 13. 1 mole of an ideal gas undergoes an isothermal expansion...
for the irreversible adiabatic heating of an ideal diatomic gas, calculate q, w, delta U, delta H, and the final temperature given p1 = 0.5 bar, p2 =3.5 bar, and T1 = 150 K
I. (30 pts.) One mole of an ideal gas with constant heat capacities and ? 5/3 is compressed adiabatically in a piston-cylinder device from T1-300 K, pi = 1 bar to p2 = 10 bar at a constant external pressure Pext"- P2 -10 bar. Calculate the final temperature, T2, and W, Q. AU, AH for this process. 2. (20 pts.) Repeat problem 1 for an adiabatic and reversible compression. 3. (20 pts.) A rigid, insulated tank is divided into two...
1.95 mol of an ideal gas with CV = 3R/2 undergoes the following transformations from an initial state T = 290 K, P = 1.000 bar. Find q, w, ∆U, ∆H and ∆S for each transformation. a) A reversible adiabatic compression until the final temperature reaches 390 K.
Consider the isothermal compression of 1 mole of a monatomic ideal gas, initially at a pressure of 0.5 bar and volume of 4 liters to a final pressure of 2 bar. Calculate the following: a. The work done if the compression is reversible-answer in Joules b. The work done if the compression is irreversible-answer in Joules
Q3. One mole of N2 in a piston-cylinder assembly undergoes an adiabatic compression from an initial state of pressure 0.5 bar and molar volume 0.05 m/mol to a final state of pressure 10 bar and molar volume 0.003 m/mol. Use the vdw eos to determine the work done on the gas. Q3. One mole of N2 in a piston-cylinder assembly undergoes an adiabatic compression from an initial state of pressure 0.5 bar and molar volume 0.05 m/mol to a final...
Please give detailed explanation for final part. Thanks. Reversible adiabatic expansion of ideal gas (This question involves working through the final section of lecture 3) Explain why the first Law for an reversible adiabatic process gives AU = -PdV, and why this equation doesn't hold for the Joule expansion. Assuming that for an ideal gas U = CVT, prove that the First Law leads to the statement that PVY is constant in a reversible adiabatic process. A container of Helium...