8) a 475 cm³ sample of gas at standard temperature and pressure is allowed to expand until it occupies a…
8) a 475 cm³ sample of gas at standard temperature and pressure is allowed to expand until it occupies a volume of 600. cm³. what temperature would be needed to return the gas to standard pressure?\n| | p | t | v | n |\n|--|--|--|--|--|\n| initial | | | | |\n| final | | | | |\n| effect | | | | |\n345k\n9. the diagram below left shows a box containing gas molecules at 25°c and 1 atm pressure. the piston is free to move.\nin the box at right, sketch the arrangement of molecules and the position of the piston at standard temperature and pressure. does the volume decrease significantly? why or why not?
Answer
Explanation:
Step1: Identify the gas - law applicable
Since the pressure is constant ($P_1 = P_2$), we use Charles's law, which is $\frac{V_1}{T_1}=\frac{V_2}{T_2}$. At standard temperature and pressure (STP), $T_1 = 273\ K$, $V_1=475\ cm^3$, and $V_2 = 600\ cm^3$.
Step2: Rearrange Charles's law to solve for $T_2$
From $\frac{V_1}{T_1}=\frac{V_2}{T_2}$, we can get $T_2=\frac{V_2T_1}{V_1}$.
Step3: Substitute the values
$T_2=\frac{600\ cm^3\times273\ K}{475\ cm^3}=\frac{163800}{475}\ K\approx345\ K$.
Answer:
$345\ K$
For the second - part:
Brief Explanations:
At standard temperature ($T = 273\ K$) compared to $25^{\circ}C(298\ K)$ with pressure constant (1 atm), according to Charles's law $\frac{V_1}{T_1}=\frac{V_2}{T_2}$. Since $T$ decreases, $V$ will decrease. The molecules will be more closely packed together as the volume of the gas decreases. The piston will move downwards. The volume decreases significantly because the temperature change from $298\ K$ to $273\ K$ is non - negligible and in the direct proportion relationship of volume and temperature at constant pressure.
Answer:
The piston will move downwards, the molecules will be more closely packed. The volume decreases significantly because of the decrease in temperature at constant pressure according to Charles's law.