If 17 L of a gas at a temperature of 67°C and a pressure of 88.89 atm has its temperature increased to 94°C and its volume decreased to 12 L, what is the new pressure?

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If 17 L of a gas at a temperature of 67°C and a pressure of 88.89 atm has its temperature increased to 94°C and its volume decreased to 12 L, what is the new pressure?

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Answer 1 · 2017-02-12T02:57:11

The new pressure will be $140 atm$ $"140 atm"$ .

Explanation:

This is an example of the combined gas law. The equation to use is:

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

where $P$ $P$ is the pressure, $V$ $V$ is the volume, and $T$ $T$ is the temperature in Kelvins. For this question, the temperature in degrees Celsius will be converted to Kelvins by adding $273.15$ $273.15$ .

Write what you know.

$P_{1} = 88.89 atm$ $P_1="88.89 atm"$
$V_{1} = 17 L$ $V_1="17 L"$
$T_{1} = 67^{\circ} C + 273.15 = 340 K$ $T_1="67"^@"C" + 273.15="340 K"$
$V_{2} = 12 L$ $V_2="12 L"$
$T_{2} = 94^{\circ} C + 273.15 = 367 K$ $T_2="94"^@"C" + 273.15="367 K"$

Write what you don't know: $P_{2}$ $P_2$

Solution
Rearrange the equation to isolate $P_{2}$ $P_2$ , substitute the known values into the equation, and solve.

$P_{2} = \frac{P_{1} V_{1} T_{2}}{V_{2} T_{1}}$ $"P_2=(P_1V_1T_2)/(V_2T_1)$

$P_{2} = \frac{88.89 atm \times 17 L \times 367 K}{12 L \times 340 K} = 140 atm$ $P_2=(88.89"atm"xx17"L"xx367"K")/(12"L"xx340"K")="140 atm"$ rounded to two significant figures

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If 17 L of a gas at a temperature of 67°C and a pressure of 88.89 atm has its temperature increased to 94°C and its volume decreased to 12 L, what is the new pressure?

1 Answer

Explanation:

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