Question

A piston-cylinder assembly contains 2 lb of air at a temperature of 540 °R and a pressure of 1 atm. The air compressed to a state where the temperature is 840 °R and the pressure is 6 atm. During the compression, there is a heat transfer from the air to the surroundings equal to 20 Btu.
1. Using the ideal gas model for air, determine the work during the process in Btu.

Answer 1

Answer:

123.9 Btu

Explanation:

The energy balance on the air is:

∆E = E2 − E1 = ∆KE + ∆PE + ∆U = Q + W

ignore  ∆KE and ∆PE,

W = ∆U − Q = m(u2 − u1) − Q;                               (u2 − u1 = 51.94 Btu/lb)

ideal gas properties is attached

W = (2 lb)(143.98 − 92.04) Btu/lb − (− 20 Btu) = 123.9 Btu

u2 − u1 ≈ cv(T2 − T1) = (0.173 Btu/lb°R)(840 − 540) °R = 51.9 Btu/lb

Answer 2

The net work done in compressing the air as given is; W = -123.8 Btu

What is the Energy Balance equation?

The equation for Energy Balance in thermodynamics is;

Q - W = ΔU

where;

Q is

ΔU is change in the internal energy of the system

Q is the net heat transfer

W is Net work done

Now, ΔU can also be written as;

ΔU = mC_v(T₂ - T₁)

C_v for air is 0.173 Btu/bm.R

We are given;

m = 2 lb

T₁ = 540 °R

T₂ = 840 °R

Q = -20 Btu (negative because heat is transferred to the surrounding)

Thus;

ΔU = 2 * 0.173 * (840 - 540)

ΔU = 103.8 Btu

Work done during the process is;

W = Q - ΔU

W = -20 - 103.8

W = -123.8 Btu

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