The first law of thermodynamics may be expressed as ∆U = (+q) + (+w) where ∆U is the increase in internal energy of the system.
- (a)State the meaning of: +q; +w.[2]
- (b)(i)The variation with pressure p of the volume V of a fixed mass of an ideal gas is shown in Fig. 2.1. [Figure 2.1] The gas undergoes a cycle of changes A to B to C to A. During the change A to B, the volume of the gas increases from 2.3 × 10⁻³ m³ to 3.8 × 10⁻³ m³. Show that the magnitude of the work done during the change A to B is 390 J.[1]
- (b)(ii)State and explain the total change, if any, in the internal energy of the gas during one complete cycle.[2]
- (c)During the change A to B, 1370 J of thermal energy is transferred to the gas. During the change B to C, no thermal energy enters or leaves the gas. The work done on the gas during this change is 550 J. Use these data and the information in (b) to complete Table 2.1. [Table 2.1][4]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [B1] +q: thermal energy transfer to system
- [B1] +w: work done on system
- [A1] (W =) 2.6×10⁵ × (3.8 – 2.3) × 10⁻³ = 390 J
- [B1] no (total) change (in internal energy)
- [B1] gas returns to its original temperature
- [B1] A to B row all correct (1370, – 390, 980)
- [B1] B to C row all correct (0, 550, 550)
- [B1] C to A row: ∆U adds to the other two ∆U values to give zero
- [B1] C to A row: w = 0 and q adds to w to give ∆U value
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