What do the left and right sides of the dome on the T-v diagram represent?

Left is saturated liquid (x = 0, v = vf); right is saturated vapour (x = 1, v = vg)

Left is saturated vapour; right is saturated liquid

Both are wet mixture boundaries at x = 0.5

What does a constant-pressure line look like inside the dome?

Horizontal at T = Tsat(P) while v goes from vf to vg

Rising steeply as the mixture heats

A curve identical to an isotherm in the superheated region

What happens at the critical point at the top of the dome?

Saturated liquid and vapour lines meet; no distinct liquid-vapour change

How does a higher-pressure isobar differ?

Higher Tsat and a shorter horizontal segment, vanishing at the critical point

Lower Tsat and a longer horizontal segment

No change since boiling is independent of pressure

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2072 Asoj · 4 marks
2071 Bhadra · 4 marks

Saturation curve on T-v diagram

Short answer

The saturation curve on a T-v diagram is a dome formed by saturated liquid (x=0) and saturated vapor (x=1) lines meeting at the critical point. Below is the wet mixture region where temperature stays constant during phase change. Left is compressed liquid, right is superheated vapor.

Explain saturation curve of two-phase mixture on T-v diagram.

A pure substance such as water, heated at constant pressure, changes from liquid to vapour at the saturation temperature of that pressure. If the states at which boiling starts and ends are joined for several pressures, we get the saturation (vapour dome) curve on the T-v diagram.

 T|          CP (critical point)
  |        .--*--.
  |      /         \
  |  liquid   wet   \  superheated
  |   | x=0  mixture x=1   vapour
  |   |  1-------2   |
  |   |  (isobar)    |
  +------------------------ v

Features

Saturated liquid line (x = 0): left part of the dome, states at which the liquid is about to boil (v = v_f).
Saturated vapour line (x = 1): right part, states at which the last liquid has vaporised (v = v_g).
Critical point: the top where the two lines meet; at P_c = 22.09 MPa, T_c = 374 °C for water, there is no distinct liquid-vapour change.
Left of x = 0: compressed (sub-cooled) liquid region.
Under the dome: wet (two-phase) mixture of liquid and vapour; T = T_sat(P) and v = v_f + x v_fg. The isobar is horizontal here.
Right of x = 1: superheated vapour region.
On a constant pressure line, T rises in the liquid, stays constant during evaporation (v_f to v_g), then rises again as vapour is superheated. Higher-pressure isobars have higher saturation temperatures and a shorter horizontal segment, which vanishes at the critical point.

✓ Answer checked by Claude Opus 5.5 (high effort)

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