A gas occupies 50 L at a pressure of 2 atm. What is the volume when the pressure is increased to 10 atm?

Question

Henry Antrim · Accepted Answer

#10# #"L"#

We're asked to calculate the new volume of a gas when its pressure is increased.

To solve this, we can use the pressure-volume relationship of gases, illustrated by Boyle's law:

#P_1V_1 = P_2V_2#

where

#P# is the initial (#"1"#) and final (#"2"#) pressure of the gas, and

#V# is the initial (#"1"#) and final (#"2"#) volume of the gas.

Let's plug in our known variables, and rearrange this equation to solve for the final volume, #V_2#:

#V_2 = (P_1V_1)/(P_2) = ((2cancel("atm"))(50"L"))/(10cancel("atm")) = color(red)(10"L"#

Thus, when the pressure of the gas was increased by a factor of #5# (#2"atm"# to #10"atm"#), the volume decreased by a factor of #5# (#50"L"# to #10"L"#).

Owen Ambrose · Answer

undefined Using Boyle's Law, we can calculate the new volume using the formula:

$P_1 \times V_1 = P_2 \times V_2$

Given $P_1 = 2 \, \text{atm}$, $V_1 = 50 \, \text{L}$, and $P_2 = 10 \, \text{atm}$, we can solve for $V_2$:

$2 \, \text{atm} \times 50 \, \text{L} = 10 \, \text{atm} \times V_2$

$100 \, \text{L} \, \text{atm} = 10 \, \text{atm} \times V_2$

$V_2 = \frac{100 \, \text{L} \, \text{atm}}{10 \, \text{atm}}$

$V_2 = 10 \, \text{L}$

So, the volume when the pressure is increased to 10 atm is 10 liters.