If $\frac{1}{3} L$ $1/3 L$ of a gas at room temperature exerts a pressure of $35 k P a$ $35 kPa$ on its container, what pressure will the gas exert if the container's volume changes to $\frac{7}{4} L$ $7/4 L$ ?

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If $\frac{1}{3} L$ $1/3 L$ of a gas at room temperature exerts a pressure of $35 k P a$ $35 kPa$ on its container, what pressure will the gas exert if the container's volume changes to $\frac{7}{4} L$ $7/4 L$ ?

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Answer 1 · 2016-02-22T22:12:12

$P = \frac{20}{3} k P a$ $P=20/3 k Pa$

Explanation:

$P_{1} \cdot V_{1} = P_{2} \cdot V_{2} Boyle-Mariotte's law$ $P_1*V_1=P_2*V_2" Boyle-Mariotte's law"$
$1/3*cancel(35)=P*cancel(7)/4$
$P=4*5/3$
$P=20/3 k Pa$

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If $\frac{1}{3} L$ $1/3 L$ of a gas at room temperature exerts a pressure of $35 k P a$ $35 kPa$ on its container, what pressure will the gas exert if the container's volume changes to $\frac{7}{4} L$ $7/4 L$ ?

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If $\frac{1}{3} L$ $1/3 L$ of a gas at room temperature exerts a pressure of $35 k P a$ $35 kPa$ on its container, what pressure will the gas exert if the container's volume changes to $\frac{7}{4} L$ $7/4 L$ ?