The pressure of 250 kPa acting on 15 cubic meters of gas at a temperature of 100 K is increased to 500 kPa and the volume is increased to 30 cubic meters. What is the temperature of the gas at this new pressure and volume?

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The pressure of 250 kPa acting on 15 cubic meters of gas at a temperature of 100 K is increased to 500 kPa and the volume is increased to 30 cubic meters. What is the temperature of the gas at this new pressure and volume?

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Answer 1 · 2016-03-11T08:56:55

The final temperature will be 400 K.

Explanation:

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

Given
Initial pressure, $P_{1} = 250 kPa$ $P_1="250 kPa"$
Initial volume, $V_{1} = {15 m}^{3}$ $V_1="15 m"^3"$
Initial temperature, $T_{1} = 100 K$ $T_1="100 K"$
Final pressure, $P_{2} = 500 kPa$ $P_2="500 kPa"$
Final volume, $V_{2} = {30 m}^{3}$ $V_2="30 m"^3"$

Unknown
Final temperature, $T_{2}$ $T_2$

Solution
Rearrange the equation to isolate $T_{2}$ $T_2$ . Substitute the given values into the equation and solve.

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

$T_{2} = \frac{T_{1} P_{2} V_{2}}{P_{1} V_{1}}$ $T_2=(T_1P_2V_2)/(P_1V_1)$

$T_2=(100"K"·500cancel"kPa"·30cancel"m"^3)/(250cancel"kPa"·15cancel"m"^3)="400K"$

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The pressure of 250 kPa acting on 15 cubic meters of gas at a temperature of 100 K is increased to 500 kPa and the volume is increased to 30 cubic meters. What is the temperature of the gas at this new pressure and volume?

1 Answer

Explanation:

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