Question

A gas at 1.8 atm, 140 K, and 4.5 L is adjusted to standard temperature and
pressure conditions. What is the new volume?

Answer 1

From the equation, $\frac{p1 * v1}{t1} = \frac{p2 * v2}{t2}$, we make v2 the subject of the formula, so that gives us: $\frac{p1 * v1 * t2}{p2 * t1}$

Standard temperature = 273k

Standard pressure = 1atm

Plugging in our values, $\frac{1.8 * 4.5 * 273}{1 * 140} = \frac{2211.3}{140} = 15.795 litres$