Question #ce4e6

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Question #ce4e6

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Answer 1 · 2016-06-10T15:27:24

The pressure would need to be increased to $16 a t m$ $16atm$

Explanation:

We can use Boyle's Law $P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ to solve for the pressure since the temperature is being held constant.

For this case we are going to use $8 L$ $8L$ to represent the original volume and $1 L$ $1L$ to represent the $\frac{1}{8}$ $1/8$ value of the final volume.

$P_{1} = 2 a t m$ $P_1 = 2atm$
$V_{1} = 8 L$ $V_1= 8L$
$P_{2} = ?$ $P_2=?$
$V_{2} = 1 L$ $V_2=1L$

$(2 a t m) (8 L) = (P_{2}) ((1 L)$ $(2atm)(8L)=(P_2)((1L)$

$\frac{(2 a t m) (8 L)}{1 L} = \frac{(P_{2}) (1 L)}{1 L}$ $((2atm)(8cancelL))/(1cancelL)=((P_2)(cancel(1L)))/(cancel(1L))$

$16 a t m = P_{2}$ $16atm = P_2$

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Question #ce4e6

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