A balloon has a volume of 0.5 L at 20°C. What will the volume be if the balloon is heated to 150°C?

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A balloon has a volume of 0.5 L at 20°C. What will the volume be if the balloon is heated to 150°C?

Assume constant pressure and mass.

2 Answers

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Answer 1 · 2016-12-26T23:05:18

Assuming pressure is constant:

Explanation:

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1=V_2/T_2$ ( $T$ $T$ in Kelvin)

$20^{o} C = 20 + 273 = 293 K$ $20^(o)C=20+273=293K$
$150^{o} C = 150 + 273 = 423 K$ $150^(o)C=150+273=423K$

$\frac{0.5}{293} = \frac{V_{2}}{423} \to V_{2} = \frac{0.5 \times 423}{293} = 0.72 L$ $0.5/293=V_2/423->V_2=(0.5xx423)/293=0.72L$

Answer 2 · 2016-12-26T23:23:06

The new volume will be $0.7 L$ $"0.7 L"$ .

Explanation:

This is an example of Charles' law, sometimes called the temperature-volume law. It states that the volume of a gas is directly proportional to the Kelvin temperature, while pressure and amount are held constant.

The equation is $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1=V_2/T_2$ , where $V$ $V$ is volume and $T$ $T$ is temperature in Kelvins.

Known
$V_{1} = 0.5 L$ $V_1="0.5 L"$
$T_{1} = 20^{\circ} C + 273.15 = 293 K$ $T_1="20"^@"C"+273.15="293 K"$
$T_{2} = 150^{\circ} C + 273.15 = 423 K$ $T_2="150"^@"C"+273.15="423 K"$

Unknown
$V_{2}$ $V_2$

Solution
Rearrange the equation to isolate $V_{2}$ $V_2$ . Substitute the known values into the equation and solve.

$V_{2} = \frac{V_{1} T_{2}}{T_{1}}$ $V_2=(V_1T_2)/T_1$

$V_{2} = \frac{(0.5 L \times 423 K)}{293 K} = 0.7 L$ $V_2=((0.5"L"xx423"K"))/(293"K")="0.7 L"$ rounded to one significant figure

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A balloon has a volume of 0.5 L at 20°C. What will the volume be if the balloon is heated to 150°C?

Assume constant pressure and mass.

2 Answers

Explanation:

Explanation:

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