At 1.01 atm and 15.0°C a constant - pressure tank has 255 $m^{3}$ $m^3$ of propane. What is the volume of the gas at 48.0°C?

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At 1.01 atm and 15.0°C a constant - pressure tank has 255 $m^{3}$ $m^3$ of propane. What is the volume of the gas at 48.0°C?

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Answer 1 · 2015-12-09T01:04:06

The volume at $321.2 K$ $"321.2 K"$ $({48.0}^{o} C)$ $("48.0"^"o""C")$ is ${251 m}^{3}$ $"251 m"^3"$ .

Explanation:

This is an example of the combined gas law, with the equation $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/(T_1)=(P_2V_2)/(T_2)$ .

Given/Known
$P_{1} = 1.01 atm$ $P_1="1.01 atm"$
$V_{1} = {225 m}^{3}$ $V_1="225 m"^3"$
$T_{1} = {15.0}^{o} C + 273.15 = 288.2 K$ $T_1="15.0"^"o""C"+273.15="288.2 K"$
$P_{2} = 1.01 atm$ $P_2="1.01 atm"$
$T_{2} = {48.0}^{o} C + 273.15 = 321.2 K$ $T_2="48.0"^"o""C"+273.15="321.2 K"$

Unknown
$V_{2}$ $V_2$

Solution
Rearrange the equation to isolate $V_{2}$ $V_2$ and solve.

$V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$ $V_2=(P_1V_1T_2)/(T_1P_2)$

$V_2=(1.01cancel"atm"xx225"m"^3xx321.2cancel"K")/(288.2cancel"K"xx1.01cancel"atm")="251 m"^3"$

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At 1.01 atm and 15.0°C a constant - pressure tank has 255 $m^{3}$ $m^3$ of propane. What is the volume of the gas at 48.0°C?

1 Answer

Explanation:

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At 1.01 atm and 15.0°C a constant - pressure tank has 255 $m^{3}$ $m^3$ of propane. What is the volume of the gas at 48.0°C?