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?

Question

Cameron Andrews · Accepted Answer

The new temperature would be 839 °C.

To solve this problem, we can use the Combined Gas Laws: #color(blue)(bar(ul(|color(white)(a/a)(p_1V_1)/T_1 = (p_2V_2)/T_2color(white)(a/a)|)))" "# 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 = "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 = "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.

Ellie Andrews · Answer

undefined To solve this, we can use the combined gas law formula:

$$ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} $$

Where:
- $ P_1 = 225 \, \text{mmHg} $
- $ V_1 = 1676 \, \text{mL} $
- $ T_1 = -85.1 \, \text{°C} + 273.15 \, \text{K} $ (Convert -85.1 °C to Kelvin)
- $ P_2 = 3.5 \, \text{atm} $
- $ V_2 = 838 \, \text{mL} $
- $ T_2 $ is what we're solving for.

First, let's convert -85.1 °C to Kelvin:
$$ T_1 = -85.1 \, \text{°C} + 273.15 \, \text{K} = 188.05 \, \text{K} $$

Now we can plug these values into the formula:
$$ \frac{225 \, \text{mmHg} \times 1676 \, \text{mL}}{188.05 \, \text{K}} = \frac{3.5 \, \text{atm} \times 838 \, \text{mL}}{T_2} $$

Now we can solve for $ T_2 $:
$$ T_2 = \frac{3.5 \, \text{atm} \times 838 \, \text{mL} \times 188.05 \, \text{K}}{225 \, \text{mmHg} \times 1676 \, \text{mL}} $$
$$ T_2 = \frac{3.5 \times 838 \times 188.05}{225 \times 1676} $$
$$ T_2 = \frac{560227.3}{377400} $$
$$ T_2 = 1.484 \, \text{K} $$

So, the temperature will be approximately 1.484 K.