Ideal Gases — Question 1

Question 1

Gauge pressure must be shifted before taking a gas-law ratio

Treat the gas as ideal, $pV=nRT$, and use $R=8.314\,\mathrm{J/(mol\,K)}$. A sealed rigid vessel has gauge pressure 200 kPa at 300 K, with ambient pressure 100 kPa. It warms to 360 K while the ambient pressure falls to 90 kPa.

1. Find final absolute and gauge pressures.

2. Find the erroneous final gauge reading obtained by multiplying the initial gauge reading by $360/300$. Quantify the error and explain its two sources.

Question 1 - Solution

Gauge pressure must be shifted before taking a gas-law ratio

(a) Method and calculation

$$p_{i,\mathrm{abs}}=200+100=300\,\mathrm{kPa},\quad\frac{p_{f,\mathrm{abs}}}{p_{i,\mathrm{abs}}}=\frac{360}{300}.$$$$p_{f,\mathrm{abs}}=360\,\mathrm{kPa},\quad p_{f,g}=360-90=270\,\mathrm{kPa}.$$

(b) Interpretation and further calculation

$$p_{f,g}^{\mathrm{wrong}}=200(360/300)=240\,\mathrm{kPa}.$$$$p_{f,g}-p_{f,g}^{\mathrm{wrong}}=100(1.2)-90=30\,\mathrm{kPa}.$$The omitted initial absolute-pressure offset contributes 20 kPa relative to unchanged ambient; the ambient fall contributes another 10 kPa.

Physical check

The equation of state uses pressure relative to vacuum and absolute temperature.