How do you calculate the initial pressure this situation: a sealed container of gas is squashed, reducing the volume from 0.5 $m^{3}$ $m^3$ to 0.25 $m^{3}$ $m^3$ , with a final pressure of 120 kPa?

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How do you calculate the initial pressure this situation: a sealed container of gas is squashed, reducing the volume from 0.5 $m^{3}$ $m^3$ to 0.25 $m^{3}$ $m^3$ , with a final pressure of 120 kPa?

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Answer 1 · 2017-09-13T07:49:23

Well, we use old Boyle's law..... $P V = k$ $PV=k$ , given constant temperature, and CONSTANT amount of gas.....and gets $P_{1} = 60 \cdot k P a$ $P_1=60*kPa$ .

Explanation:

And since $P V = k$ $PV=k$ , and thus for a PISTON at constant temperature, we write.....

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ .............

And we solve for $P_{1} = \frac{120 \cdot k P a \times 0.25 \cdot m^{3}}{0.5 \cdot m^{3}} = ? ? \cdot k P a$ $P_1=(120*kPaxx0.25*m^3)/(0.5*m^3)=??*kPa$ .

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How do you calculate the initial pressure this situation: a sealed container of gas is squashed, reducing the volume from 0.5 $m^{3}$ $m^3$ to 0.25 $m^{3}$ $m^3$ , with a final pressure of 120 kPa?

1 Answer

Explanation:

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How do you calculate the initial pressure this situation: a sealed container of gas is squashed, reducing the volume from 0.5 m3m^3 to 0.25 m3m^3, with a final pressure of 120 kPa?

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Explanation:

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How do you calculate the initial pressure this situation: a sealed container of gas is squashed, reducing the volume from 0.5 $m^{3}$ $m^3$ to 0.25 $m^{3}$ $m^3$ , with a final pressure of 120 kPa?