Sol. The physical system consists of a certain mass m (which can easily be calculated) of an ideal gas whose molecular mass M is known.

The initial macro-state of the system (point 1 in Figure) is known (the standard pressure p 0 ≈ 10 5 Pa, the standard temperature T 0 = 273 K, and the volume V = 1 m 3 specified in the terms of the problem). The states and processes in which the system takes part are depicted on a p-V diagram (see Figure). Let us find the parameters of the second (point 2) and third (point 3) macro-state of the system. For this we use the ideal gas law (Figure) and the definitions of isoprocesses:

p 2 = np 0 , V 2 = V 0 ,

T 2 = = …..(i)

p 3 = p 2 = np 0 , V 3 = kV 0 ,

T 3 = = …..(ii)

Since the processes in which the system participates are quasi-static and polytropic, the sought quantities can be found by using the formulas. The variation in internal energy is

Δ U = (T 3 – T 0 ) =

= p 0 V 0 (nk – 1). …..(iii)

In an isochoric process dV = 0 and no work is done. The work done in the isobaric process is

W = p 2 (V 3 – V 2 ) = np 0 (kV 0 – V 0 ) = p 0 V 0 n (k – 1)...(iv)

The amount of heat

Q = Q 1 + Q 2 = C v (T 2 – T 0 ) + C p (T 3 – T 2 )

= [i(nk – 1) + 2n(k – 1)], …..(v)

Where we have allowed for Mayer's formula

C p = C v + R. ……(vi)

The amount of heat can be calculated by using the first law of thermodynamics:

Q = Δ U + W = p 0 V 0 (nk – 1) + p 0 V 0 n(k – 1)

= [i(nk – 1) + 2n (k – 1)];

this coincides with the earlier result (v). To carry out a numerical calculation we must select reasonable values for n and k. Equations (ii) show that the value of n determines the maximum value of pressure, p 3 = np 0 , while the product nk determines the maximum temperature, T 3 = nkT 0 . The value of n cannot exceed n max = 100, since at pressures equal to (or greater than) 100 p 0 the gas ceases to be ideal. The product nk cannot exceed the value (nk) max ≈ 10 since at temperature T 3 ≈ 10 T 0 ≈ 3 × 10 3 K (and higher) the walls of the vessel containing the hydrogen could melt (they must be cooled) and molecular hydrogen transforms into atomic hydrogen; at still higher temperatures the atomic hydrogen transforms into hydrogen plasma. If we put n = 5 and k = 2, we find that Eqs. (iii), (iv), and (v) yield

Δ U ≈ 2.2 × 10 6 J, W = 5 × 10 5 J, Q ≈ 2.7 × 10 6 J.

It is advisable to study the following problem: what must be the relationship between n and k (at nk = const) if we want the ratio W/Q to be maximal? But we will leave this problem for the reader to solve. Instead we pose the following problem: how do the sought quantities change if the system proceeds from the initial state to the final state quasi-statically along the dashed straight line in Figure? The answer can easily be found if we study for the first law of thermodynamics. The variation of internal energy Δ U, does not depend on the type of process but on the initial and final states of the system. Hence, the variation in internal energy does not change: Δ U ≈ 2.2 × 10 6 J. The amount of work done by the system will decrease (the area of the trapezoid ABDE is smaller than the area of the rectangle ACDE). Hence, according to the first law of thermodynamics, the system will receive less heat.