Under a pressure of $1.2 \cdot a t m$ $1.2atm$ a gas occupies an $8 \cdot m L$ $8mL$ volume in a piston. What pressure develops if the volume is increased to $12 \cdot m L$ $12*mL$ ?

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Under a pressure of $1.2 \cdot a t m$ $1.2atm$ a gas occupies an $8 \cdot m L$ $8mL$ volume in a piston. What pressure develops if the volume is increased to $12 \cdot m L$ $12*mL$ ?

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Answer 1 · 2017-01-16T21:22:43

$P_{2} = 0.8 \cdot a t m$ $P_2=0.8*atm$

Explanation:

Boyle's Law holds that $P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ at constant temperature.

And thus $P_{2} = \frac{P_{1} V_{1}}{V_{2}} = \frac{1.2 \cdot a t m \times 8 \cdot m L}{12 \cdot m L} = 0.8 \cdot a t m$ $P_2=(P_1V_1)/V_2=(1.2*atmxx8*mL)/(12*mL)=0.8*atm$ .

This decrease in pressure is reasonable, given that we have INCREASED the volume.

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Under a pressure of $1.2 \cdot a t m$ $1.2atm$ a gas occupies an $8 \cdot m L$ $8mL$ volume in a piston. What pressure develops if the volume is increased to $12 \cdot m L$ $12*mL$ ?

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Explanation:

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Under a pressure of 1.2⋅atm1.2*atm a gas occupies an 8⋅mL8*mL volume in a piston. What pressure develops if the volume is increased to 12⋅mL12*mL?

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Under a pressure of $1.2 \cdot a t m$ $1.2atm$ a gas occupies an $8 \cdot m L$ $8mL$ volume in a piston. What pressure develops if the volume is increased to $12 \cdot m L$ $12*mL$ ?