Answer:
"457.2 mm.Hg" is the right solution.
Explanation:
Given:
Pressure,
[tex]P_1=600 \ mm.Hg[/tex]
[tex]P_2 = ?\\[/tex]
Volume,
[tex]V_1=70.0 \ L[/tex]
[tex]V_2=90.0 \ L[/tex]
Temperature,
[tex]T_1=20^{\circ}[/tex]
or,
[tex]=293 \ K[/tex]
[tex]T_2=15.0^{\circ}[/tex]
or,
[tex]=288 \ K[/tex]
As we know,
⇒ [tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]
By putting all the given values in the above expression, we get
⇒ [tex]\frac{0.789\times 70}{293} =\frac{P_2\times 90}{288}[/tex]
⇒ [tex]0.188=\frac{P_2\times 90}{288}[/tex]
By applying cross-multiplication, we get
⇒ [tex]P_2=\frac{0.188\times 288}{90}[/tex]
⇒ [tex]=\frac{54.144}{90}[/tex]
⇒ [tex]=0.6016 \ atm[/tex]
or,
⇒ [tex]=457.2 \ mm.Hg[/tex]
