Question
Question: An open vessel at \({{27}^{\circ }}C\) is heated until three-fifth of the air in it has been expelle...
An open vessel at 27∘C is heated until three-fifth of the air in it has been expelled. Assuming that the volume of the vessel remains constant, find the temperature to which the vessel has been heated?
Solution
In order to solve this question, we must look at the parameters given to us in the question. We should relate the temperature and number of moles by using the ideal gas equation, as volume (V), pressure (P) and real gas constant (R) remain constant.
Complete Solution :
- As we are provided with the information that, there is an open vessel with a temperature of 27∘C. So,
27∘C=273.15+27=300K
- We can say that there is a change in moles of the gas, as three-fifth of the air escapes. Hence, let the initial moles in air be n1. And n2 be the number of moles after three fifths of gas escapes.
n1=1mole
n2=1−53=52 moles.
- Ideal gas equation is PV= nRT
As pressure, volume and real gas constant are constant, we can write above equation as:
nT=constant
Therefore, n1T1=n2T2
n2n1=T1T2
- Now, by putting values in above equation we get:
25=300T2
T2=750K
- Hence, we can conclude that the temperature to which the vessel has been heated is 750K.
Note: - As we know that ideal gas equation is an approximation of the behaviour of gases under ideal conditions, and is the equation of state of an ideal gas (hypothetical).
- It is basically combination of empirical laws like Charles law, Boyle’s law, Avogadro’s law and Gay-Lussac’s law.