# A "2270 mL" sample of gas has a pressure of "570. mmHg" at "25"^@"C". What is the pressure when the volume is decreased to "1250 mL" and the temperature is increased to "175"^@"C"?

Jan 12, 2016

1556 mm Hg

#### Explanation:

${V}_{1} / {V}_{2} = \frac{{T}_{1} {P}_{2}}{{T}_{2} {P}_{1}}$

T_1 = (25 +273.15) K = 298.15 K; T_2 = (175 +273.15) K = 448.15 K

$\implies {P}_{2} = \frac{{V}_{1} {T}_{2} {P}_{1}}{{V}_{2} {T}_{1}}$

$\implies {P}_{2} = \left(2270 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL")))*448.15 color(red)(cancel(color(black)("K")))×"570. mm Hg")/ (1250color(red)(cancel(color(black)("mL")))*298.15 color(red)(cancel(color(black)("K}}}}\right)$

$\implies {P}_{2} = \text{1556 mm Hg}$

Jan 13, 2016

The final pressure is 1560 mmHg.

#### Explanation:

Use the combined gas law, which relates pressure, volume, and temperature.

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{{T}_{2}}$

Note: Temperature must be converted from degrees Celsius to Kelvins.

Known/Given
${P}_{1} = \text{570. mmHg}$
${V}_{1} = \text{2270 mL}$
${T}_{1} = \text{25"^"o""C"+273.15="298 K}$
${V}_{2} = \text{1250 mL}$
${T}_{2} = \text{175"^"o""C"+273.15="448 K}$

Unknown
${P}_{2} = \text{???}$

Solution
Rearrange the equation to isolate ${P}_{2}$ and solve.
$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{{T}_{2}}$

${P}_{2} = \frac{{P}_{1} {V}_{1} {T}_{2}}{{T}_{1} {V}_{2}}$

P_2=(570."mmHg"xx2270cancel"mL"xx448cancel"K")/(298cancel"K"xx1250cancel"mL")="1560 mmHg" (rounded to three significant figures)