A constant temperature, a $12 \cdot L$ $12L$ volume of gas in a piston, is compressed to $3 \cdot L$ $3L$ . If the original pressure was $41 \cdot k P a$ $41*kPa$ , what is the final pressure?

Question

A constant temperature, a $12 \cdot L$ $12L$ volume of gas in a piston, is compressed to $3 \cdot L$ $3L$ . If the original pressure was $41 \cdot k P a$ $41*kPa$ , what is the final pressure?

2 Answers

Impact of this question

1838 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-11T18:04:03

The pressure is $= 164 k P a$ $=164kPa$

Explanation:

We apply Boyle's Law

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ , the temperature is constant

The initial pressure is $P_{1} = 41 k P a$ $P_1=41kPa$

The initial volume is $V_{1} = 12 L$ $V_1=12L$

The final volume is $V_{2} = 3 L$ $V_2=3L$

The final pressure is

$P_{2} = \frac{V_{1}}{V_{2}} \cdot P_{1} = \frac{12}{3} \cdot 41 = 164 k P a$ $P_2=V_1/V_2*P_1=12/3*41=164kPa$

Answer 2 · 2017-08-11T18:04:51

Well, we use old Boyle's law.........and get $P_{2} = 164 \cdot k P a$ $P_2=164*kPa$ .

Explanation:

At constant temperature, and constant amount of gas, the product $P \times V = k$ $PxxV=k$ , where $k$ $k$ is some constant....

And thus, if we solve for $k$ $k$ at different conditions of volume and pressure, then $P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ .

The utility of this equation is that we can use whatever whack units of volume and pressure we like, $pints, pounds per square inch, torr, atmospheres, quarts$ $"pints, pounds per square inch, torr, atmospheres, quarts"$ , as long as we are consistent.

And so... $P_{2} = \frac{P_{1} V_{1}}{V_{2}} = \frac{41 \cdot k P a \times 12 \cdot L}{3 \cdot L} = 164 \cdot k P a$ $P_2=(P_1V_1)/V_2=(41*kPaxx12*L)/(3*L)=164*kPa$

Science

Math

Humanities

... and beyond

A constant temperature, a $12 \cdot L$ $12L$ volume of gas in a piston, is compressed to $3 \cdot L$ $3L$ . If the original pressure was $41 \cdot k P a$ $41*kPa$ , what is the final pressure?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A constant temperature, a 12⋅L12*L volume of gas in a piston, is compressed to 3⋅L3*L. If the original pressure was 41⋅kPa41*kPa, what is the final pressure?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

A constant temperature, a $12 \cdot L$ $12L$ volume of gas in a piston, is compressed to $3 \cdot L$ $3L$ . If the original pressure was $41 \cdot k P a$ $41*kPa$ , what is the final pressure?