A compressor fills a 12.0 L scuba cylinder until the pressure is 20.0 MPa…

Step 1. Moles. $n = \dfrac{pV}{RT} = \dfrac{20.0 \times 10^6\ \text{Pa} \times 0.0120\ \text{m}^3}{8.31\ \text{J/(mol K)} \times 330\ \text{K}} = 87.5\ \text{mol}$.

Step 2. After cooling. $p_2 = 20.0\ \text{MPa} \times \dfrac{295\ \text{K}}{330\ \text{K}} = 17.9$ MPa.

Step 3. Open air. We want $V$, so $V = \dfrac{nRT}{p}$. The top line is $nRT = 87.5\ \text{mol} \times 8.31\ \text{J/(mol K)} \times 295\ \text{K} = 2.145 \times 10^5\ \text{J}$. So $V = \dfrac{2.145 \times 10^5\ \text{J}}{100 \times 10^3\ \text{Pa}} = 2.145\ \text{m}^3 = 2145$ L.