Question

What pressure is required to reduce 77 mL of a gas at standard conditions to 16 mL at a temperature of 24° C? Answer in units of atm.

Answer 1

We want to find the P2, given

V1 = 77 mL

V2 = 16 mL

T2 = 24 degrees celsius = 24 + 273 = 297 K

At standard conditions we know that temperature = 0 degrees celsius and pressure = 1 atm so:

T1 = o degrees celsius = 0 + 273 = 273 K

P1 = 1 atm

P2 = ?

To solve this problem, we will use the following equation:

$\begin{gathered} \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \\ \end{gathered}$

P2 = (1 atm x 77 mL x 297 K)/(273 K x 16 mL)

P2 = 5.235576923 atm