A compressor pumps $70 L$ of air into a $6 L$ tank with the temperature remaining unchanged. If all the air is originally at $1 a t m$ . What is the final absolute pressure of the air in the tank?

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Solution

we use Boyle's law, since the temperature is the constant.
$p_{0} V_{0} = p V$

Here,
$p_{0} = 1 a t m$
$V = 6 L$

$\begin{matrix} V_{0} = 70 + 6 = 76 L 1 (76) = p (6) p = 12.7 a t m \end{matrix}$

Hence, $12.7 a t m$ Absolute pressure.

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A compressor pumps 70 L of air into a 6 L tank with the temperature remaining unchanged. If all the air is originally at 1 atm. What is the final absolute pressure of the air in the tank?

we use Boyle's law, since the temperature is the constant. p0V0=pVHere, p0=1atmV=6LV0=70+6=76L1(76)=p(6)p=12.7atm Hence, 12.7atm Absolute pressure.

A compressor pumps $70 L$ of air into a $6 L$ tank with the temperature remaining unchanged. If all the air is originally at $1 a t m$ . What is the final absolute pressure of the air in the tank?

we use Boyle's law, since the temperature is the constant.
$p_{0} V_{0} = p V$

Here,
$p_{0} = 1 a t m$
$V = 6 L$

$\begin{matrix} V_{0} = 70 + 6 = 76 L 1 (76) = p (6) p = 12.7 a t m \end{matrix}$

Hence, $12.7 a t m$ Absolute pressure.