A gas occupies 1676 mL at -85.1 °C and 225 mmHg. What will its temperature be when its volume is 838 mL at 3.5 atm?

May 9, 2017

The new temperature would be 839 °C.

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \frac{{p}_{1} {V}_{1}}{T} _ 1 = \frac{{p}_{2} {V}_{2}}{T} _ 2 \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can solve this equation for ${T}_{2}$.

T_2 = T_1 × p_2/p_1 × V_2/V_1

In this problem,

p_1 = 225 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.2960 atm"; p_2 = "3.5 atm"

${V}_{1} = \text{1676 mL";color(white)(mmmmmmmmmmmmmmmll) V_2 = "838 mL}$

T_1 = "(-85.1 + 273.15) K = 188.05 K";color(white)(mmmmm) T_2 = ?

${T}_{2} = \text{188.05 K" × (3.5 color(red)(cancel(color(black)("atm"))))/(0.2960 color(red)(cancel(color(black)("atm")))) × (838 color(red)(cancel(color(black)("mL"))))/(1676 color(red)(cancel(color(black)("mL")))) = "1112 K" = "839 °C}$

The new temperature is 839 °C.