$2.50 L$ $"2.50 L"$ 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?

1 Answer

Impact of this question

2374 views around the world

You can reuse this answer
Creative Commons License

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.)

Science

Math

Humanities

... and beyond

$2.50 L$ $"2.50 L"$ 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:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

"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:

Related questions

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?