Answer:

Explanation:
Here, we want to get the volume of the nitrogen gas at the lower temperature
From Charles' law, we know that volume and temperature (in Kelvin) are directly proportion
The mathematical relationship is:

Where:
V1 is the initial volume which is 410 L
V2 is the final volume which is unknown
T1 is the initial temperature which we will convert to Kelvin by adding 273.15 K, we have it as 61 + 273.15 = 334.15 K
T2 is the final temperature which we have to convert to Kelvin by adding it to 273.15K : We have that as -25 + 273.15 = 248.15 K
Substituting the values, we have it that:
Explanation:
vraiLe point d'ébullition est la température à laquelle une substance passe de la phase liquide à la phase gazeuse
Answer:
Option A = 2.2 L
Explanation:
Given data:
volume of one mole of gas = 22.4 L
Volume of 0.1 mole of gas at same condition = ?
Solution:
It is known that one mole of gas at STP occupy 22.4 L volume. The standard temperature is 273.15 K and standard pressure is 1 atm.
For 0.1 mole of methane.
0.1/1 × 22.4 = 2.24 L
0.1 mole of methane occupy 2.24 L volume.
As the volume of the confined gas decreases at the constant temperature, the pressure exerted by the gas increases.