Question

1. a gas that has a volume of 26 l, a temperature of 50°c, and an unknown pressure has its volume increased to 33 l and its temperature decreased to 37°c. if i measure the pressure after the change to be 2.1 atm, what was the original pressure of the gas?

p1 =
v1 =
t1 =
p2 =
v2 =
t2 =

Explanation:

Step1: Convert temperaturas a Kelvin

$T_1 = 50 + 273 = 323\ K$
$T_2 = 37+ 273 = 310\ K$

Step2: Identificar valores de volumen y presión

$V_1 = 26\ L$, $V_2 = 33\ L$, $P_2 = 2.1\ atm$

Step3: Aplicar ley de los gases ideales combinada

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$
$P_1=\frac{P_2V_2T_1}{V_1T_2}$

Step4: Sustituir valores y calcular

$P_1=\frac{2.1\times33\times323}{26\times310}$
$P_1=\frac{22290.9}{8060}$
$P_1 = 2.77\ atm$

Answer:

$2.77\ atm$