How to find the initial pressure?

enter image source here
Here's my work :
#T_1=74+273.15 = 347.15 k #
#T_2 = 120+273.15 = 393.15 k #
#P_2=1.79 atm #
#P_1 T_1 = P_2 T_2#
#P_1 = ( 393.15*1.79)/347.15 = 2.027 atm#
Could somebody solve this problem ?

1 Answer
Jan 8, 2018

Answer:

The initial pressure was 1.58 atm.

Explanation:

Since the volume is constant but the pressure and temperature are changing, this is an example of Gay-Lussac's Law:

#color(blue)(bar(ul(|color(white)(a/a)p_1/T_1=p_2/T_2color(white)(a/a)|)))" "#

We can rearrange the above formula to get

#p_1=p_2 × T_1/T_2#

Your data are:

#p_1 = "?";color(white)(mmmm)T_1 =color(white)(ll) "(74 + 273.15) K"color(white)(l) = "347.15 K"#
#p_2 = "1.79 atm"; color(white)(l)T_2 = color(white)(l)"(120 + 273.15) K" = "393.15 K"#

#p_1 = "1.79 atm" × (347.15 color(red)(cancel(color(black)("K"))))/(393.15 color(red)(cancel(color(black)("K")))) = "1.58 atm"#

The initial pressure was 1.58 atm.

(You used the wrong formula for Gay-Lussac's Law.)