A 350 mL sample of air collected at 35°C has a pressure of 550 torr. What pressure will the air exert if it is allowed to expand to 425 mL at 57°C?

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Answer 1 · 2017-06-17T22:08:55

$485$ $485$ $torr$ $"torr"$

We can solve this equation using the combined gas law:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/(T_1) = (P_2V_2)/(T_2)$

Remember, always use the absolute (Kelvin) temperature when working with gas equations.

The temperature conversions are

$T_{1} =^{o} C + 273 = 35^{o} + 273 = 308$ $T_1 = ""^"o""C" + 273 = 35^"o" + 273 = color(red)(308$ $K$ $color(red)("K"$

$T_{2} = 57^{o} C + 273 = 330$ $T_2 = 57^"o""C" + 273 = color(green)(330$ $K$ $color(green)("K"$

Since we're trying to find the final pressure, let's rearrange this equation to solve for $P_{2}$ $P_2$ :

$P_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} V_{2}}$ $P_2 = (P_1V_1T_2)/(T_1V_2)$

Our known values:

$P_{1} = 550$ $P_1 = 550$ $torr$ $"torr"$

$V_{1} = 350$ $V_1 = 350$ $mL$ $"mL"$

$T_{1} = 308$ $T_1 = color(red)(308$ $K$ $color(red)("K"$

$V_{2} = 425$ $V_2 = 425$ " $m L$ $mL"$

$T_{2} = 330$ $T_2 = color(green)(330$ $K$ $color(green)("K"$

Let's plug these into the equation to find the final pressure:

$P_2 = ((550"torr")(350cancel("mL"))(330cancel("K")))/((308cancel("K"))(425cancel("mL"))) = color(blue)(485$ $color(blue)("torr"$

The final pressure after it is subjected to these changes is thus $color(blue)(485$ $sfcolor(blue)("torr"$ .