Unit 01 · Lesson 05 of 08

The zero on a gauge is a choice.

A pressure gauge can read zero in a pressurized atmosphere. A temperature of 20°C is not twice as hot as 10°C. Reference points matter whenever a measurement enters an equation.

Read the scales correctly

What you’ll be able to do

Convert gauge and absolute pressure.
Use absolute temperature in gas-law ratios.
Distinguish a temperature value from a temperature difference.

Before you touch the controls

If a gauge $reads -20 \mathrm{kPa}$ , is the absolute pressure negative?

Explore the modelChange a control. Follow the response.

Absolute pressure150 kPa

Gauge +50 kPa + atmosphere 100 kPa.

Gauge pressure50 kPaLocal atmosphere100 kPa

Pressure is force per area.

Pressure describes normal compressive force per unit area. A pascal is a newton per square meter. A pressure difference across a piston or surface creates a net force. Real force balances may also include weight, friction, or springs.

F=\Delta p\,A\quad\text{(uniform pressure difference)}

Gauge pressure subtracts a reference.

Absolute pressure is measured from vacuum. Gauge pressure is usually measured relative to local atmospheric pressure. A negative gauge pressure means below that reference, not below absolute zero pressure. Atmospheric pressure is not always exactly 101.3 kPa.

p_{abs}=p_g+p_{atm}

Use kelvin for temperature ratios.

Kelvin starts at absolute zero. Celsius differs by a constant offset: $T[\mathrm K]=T[{}^\circ\mathrm C]+273.15$ . A change of 1°C is a change of 1 K, but ratios of Celsius readings do not describe ratios of absolute temperatures. Ideal-gas relations require absolute pressure and absolute temperature.

A worked example

A sub-atmospheric vessel.

A vessel $reads -30 \mathrm{kPa}$ gauge while the local atmosphere is 100 kPa.
Absolute $pressure = -30 + 100 = 70 \mathrm{kPa}$ . It is above vacuum.
If the outside exceeds the inside by 30 kPa across 0.01 m², the resulting inward pressure force is 300 N.

Make the idea your own

Predict. Calculate. Question the model.

Try each problem before opening its solution. Then change a value to test whether the reasoning still holds.

01 · Calculate

Read a vacuum gauge.

A vessel has gauge $pressure -25 \mathrm{kPa}$ at an atmospheric pressure of 98 kPa. Find its absolute pressure.

A small hint

Add the local atmospheric reference, even when the gauge value is negative.

Explore the visual solution

Absolute vessel pressure73 kPa

-25 kPa $gauge + 98 \mathrm{kPa}$ atmosphere. Bars share the absolute-pressure reference.

What if you change it?Gauge pressure-25 kPa

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Use $p_{abs}=p_g+p_{atm}$ .

Carry this forward. An ordinary absolute-pressure measurement cannot be negative.

02 · Reason

Twice the Celsius reading?

A gas in a rigid sealed vessel warms from 20°C to 40°C. Treat it as ideal. Does its absolute pressure double?

A small hint

For fixed mass and volume, $\frac{p_2}{p_1}=\frac{T_2}{T_1}$ with temperatures in kelvin.

Explore the visual solution

Initial temperature293.15 K

Final temperature313.15 K

Final / initial absolute pressure1.0682 ×

The fractional pressure increase is 6.82%. Fixed mass, fixed volume, ideal gas.

What if you change it?Final temperature40 °C

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Convert: $T_1 = 293.15 \mathrm{K}$ and $T_2 = 313.15 \mathrm{K}$ .

Carry this forward. A 20°C change equals a 20 K change, but 40°C/20°C is not a valid gas-law temperature ratio.

03 · Challenge

Can a loaded piston stay in balance?

A frictionless horizontal piston has area 0.010 m². Gas on one side is at 250 kPa absolute and the atmosphere on the other is at 100 kPa. What additional opposing force holds it stationary?

A small hint

Atmospheric pressure already supplies part of the opposing force.

Explore the visual solution

Additional opposing force1,500 N

( $250 - 100$ ) $\mathrm{kPa} \times 1,000$ $\mathrm{Pa}/\mathrm{kPa} \times 0.010$ m².

What if you change it?Gas absolute pressure250 kPa

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Net pressure $difference = 250 - 100 = 150 \mathrm{kPa}$ .

Carry this forward. Use the pressure difference when calculating the net force across a surface.

Your learning path

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