Question

If 10.5 L of a gas at 0.98 atm has its pressure increased to 1.50 atm, what is the new volume?

Answer 1

Answer:

The new volume of this gas is 6.86 liters.

Assumption: the temperature of this gas stays the same, and this gas is ideal such that Boyle's Law applies.

Explanation:

By Boyle's Law, the volume of an ideal gas shall be inversely proportional to the pressure on it when temperature stays the same (as in an isothermal process.)  

In other words,

$\displaystyle V \propto \frac{1}{P}$,

where

• $V$ is the volume of the gas, and
• $P$ is the pressure on the gas.

$P_1 \cdot V_1 = P_2 \cdot V_2$.

$\displaystyle V_2 = \frac{P_1 \cdot V_1 }{P_2}$.

Assume that this gas is ideal. Also assume that this increase in pressure is isothermal. Apply Boyle's Law to find the new volume of this gas:

$\displaystyle V_2 = \frac{P_1 \cdot V_1 }{P_2} = \rm \frac{0.98\;atm \times 10.5\; L}{1.50\; atm} = 6.86\; L$.