Question

Please?!?! If I have 17 L of gas at a tempture of 340.15 K and a pressure of 88.89 Pa what will be the presure of the gas if I raise the temperature to 367.15 K and decrease the volume to 12L

Answer 1

Answer:

135.92Pa

Explanation:

This question requires using the Combined Gas Law, $\frac{P_{1}V_{1} }{T_{1}} =\frac{P_{2}V_{2} }{T_{2}}$, wherein the P1, V1, and T1 are the initial gas conditions and P2, V2, and T2 are conditions after. We are given all three of the initial conditions and two out of the three conditions after the changes, which means we can solve for the missing variable in the equation.

Here are the steps for solving the equation:

$\frac{88.89*17 }{340.15} =\frac{P_{2}*12 }{367.15}$

multiply the numerator of the initial side of the equation

$\frac{1511.13}{340.15} = \frac{P_{2}*12}{367.15}$

divide 1511.13 by 340.15

$4.44254 = \frac{P_{2}*12}{367.15}$

multiply both sides of the equation by 367.15

$1631.078582 = P_{2}*12$

divide both sides by 12

$135.9232151 = P_{2}$

If you round this to the nearest hundreds place you get a pressure of  135.92Pa