What is the initial pressure of a gas having an initial temperature of 90.5K, an initial volume of 40.3 L, final pressure of .83 atm, a final temperature of .54 K and a final volume of 2.7L?

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What is the initial pressure of a gas having an initial temperature of 90.5K, an initial volume of 40.3 L, final pressure of .83 atm, a final temperature of .54 K and a final volume of 2.7L?

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Answer 1 · 2015-12-05T07:05:30

The initial volume is $9.3 atm$ $"9.3 atm"$ .

Explanation:

Use the combined gas law. $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$

Given
$V_{1} = 40.3 L$ $V_1="40.3 L"$
$T_{1} = 90.5 K$ $T_1="90.5 K"$
$P_{2} = 0.83 atm$ $P_2="0.83 atm"$
$V_{2} = 2.7 L$ $V_2="2.7 L"$
$T_{2} = 0.54 K$ $T_2="0.54 K"$

Unknown
$P_{1}$ $P_1$

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

Solution
Rearrange the equation to isolate $P_{1}$ $P_1$ and solve.

$P_{1} = \frac{P_{2} V_{2} T_{1}}{T_{2} V_{1}}$ $P_1=(P_2V_2T_1)/(T_2V_1)$

$P_1=((0.83"atm" xx2.7cancel"L"xx90.5cancel"K"))/((0.54cancel"K"xx40.3cancel"L"))="9.3 atm"$

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What is the initial pressure of a gas having an initial temperature of 90.5K, an initial volume of 40.3 L, final pressure of .83 atm, a final temperature of .54 K and a final volume of 2.7L?

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Explanation:

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