Question

A gas sample is heated from -20.0 C to 57.0 C and the volume is increased from 2.00 L to 4.50 L If the initial pressure is .140 atm, what’s the final pressure?

Answer 1

Answer:

0.081atm

Explanations:

According to general gas equation expressed as:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where:

• P1 and P2 are the ,initial and final pressure

,

• V1 and V2 are the ,initial and final volume

,

• T1 and T2 are the i,nitial and final temperature

Given the following parameters

$\begin{gathered} V_1=2.00L \\ V_2=4.50L \\ T_1=-20.0^0C+273=253K \\ T_2=57^0C+273=330K \\ P_1=0.140atm \end{gathered}$

Required

Final Pressure P2

Substitute the given parameters into the formula

$P_2=\frac{P_1V_1T_2}{V_2T_1}$

Substitute the given following parameters into the formula

$\begin{gathered} P_2=\frac{0.14\times2.00L\times330K}{4.50L\times253K} \\ P_2=\frac{92.4}{1138.5} \\ P_2=0.081atm \end{gathered}$

Hence the final pressure of the gas sample is 0.081atm