A piston of $11 \cdot L$ $11L$ volume under $42 \cdot k P a$ $42kPa$ pressure is compressed to a volume of $8 \cdot L$ $8*L$ ; what is the new pressure?

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A piston of $11 \cdot L$ $11L$ volume under $42 \cdot k P a$ $42kPa$ pressure is compressed to a volume of $8 \cdot L$ $8*L$ ; what is the new pressure?

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Answer 1 · 2016-08-03T17:24:06

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ . $P_{2} ≅ 58 \cdot k P a$ $P_2~=58*kPa$

Explanation:

$P_{2} = \frac{P_{1} V_{1}}{V_{2}}$ $P_2=(P_1V_1)/V_2$ $=$ $=$ $\frac{42 \cdot k P a \times 11 \cdot L}{8 \cdot L}$ $(42*kPaxx11*L)/(8*L)$ $=$ $=$ $? ?$ $??$

This is not a very realistically described experiment. How does the volume decrease? Was the gas originally in a piston whose volume reduced? Well, the reaction does not say so, but the volume reduced somehow.

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A piston of $11 \cdot L$ $11L$ volume under $42 \cdot k P a$ $42kPa$ pressure is compressed to a volume of $8 \cdot L$ $8*L$ ; what is the new pressure?

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A piston of 11⋅L11*L volume under 42⋅kPa42*kPa pressure is compressed to a volume of 8⋅L8*L; what is the new pressure?

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Explanation:

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A piston of $11 \cdot L$ $11L$ volume under $42 \cdot k P a$ $42kPa$ pressure is compressed to a volume of $8 \cdot L$ $8*L$ ; what is the new pressure?