$2.50 L$ $"2.50 L"$ of a gas at $303 K$ $"303 K"$ and $100 kPa$ $"100 kPa"$ are heated. The new volume is $3.75 L$ $"3.75 L"$ and the new pressure is $90.0 kPa$ $"90.0 kPa"$ . What is the new temperature?

Question

$2.50 L$ $"2.50 L"$ of a gas at $303 K$ $"303 K"$ and $100 kPa$ $"100 kPa"$ are heated. The new volume is $3.75 L$ $"3.75 L"$ and the new pressure is $90.0 kPa$ $"90.0 kPa"$ . What is the new temperature?

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Answer 1 · 2017-08-09T04:17:47

$409 K$ $"409 K"$ to three significant figures.

Explanation:

This question is an example of the combined gas law. The formula is:

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

Organize the data:

Known

$P_{1} = 100 kPa$ $P_1="100 kPa"$

$V_{1} = 2.50 L$ $V_1="2.50 L"$

$T_{1} = 303 K$ $T_1="303 K"$

$P_{2} = 90.0 kPa$ $P_2="90.0 kPa"$

$V_{2} = 3.75 L$ $V_2="3.75 L"$

Unknown

$T_{2}$ $T_2$

Solution

Rearrange the formula to isolate $T_{2}$ $T_2$ . Insert your data and solve.

$T_{2} = \frac{P_{2} V_{2} T_{1}}{P_{1} V_{1}}$ $T_2=(P_2V_2T_1)/(P_1V_1)$

$T_2=((90.0color(red)cancel(color(black)("kPa"))xx3.75color(red)cancel(color(black)("L"))xx303"K"))/((100color(red)cancel(color(black)("kPa"))xx2.50color(red)cancel(color(black)("L"))))="409K"$ rounded to three significant figures

( $"100 kPa"$ is an exact number so it has an infinite number of significant figures.)

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$2.50 L$ $"2.50 L"$ of a gas at $303 K$ $"303 K"$ and $100 kPa$ $"100 kPa"$ are heated. The new volume is $3.75 L$ $"3.75 L"$ and the new pressure is $90.0 kPa$ $"90.0 kPa"$ . What is the new temperature?

1 Answer

Explanation:

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"2.50 L"2.50 L"2.50 L" of a gas at "303 K"303 K"303 K" and "100 kPa"100 kPa"100 kPa" are heated. The new volume is "3.75 L"3.75 L"3.75 L" and the new pressure is "90.0 kPa"90.0 kPa"90.0 kPa". What is the new temperature?

1 Answer

Explanation:

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Impact of this question

$2.50 L$ $"2.50 L"$ of a gas at $303 K$ $"303 K"$ and $100 kPa$ $"100 kPa"$ are heated. The new volume is $3.75 L$ $"3.75 L"$ and the new pressure is $90.0 kPa$ $"90.0 kPa"$ . What is the new temperature?