Sigma Percentile
JEE Main 2019
LEVELJEE Main

Animated Solution for Chemistry - States of Matter: An open vessel at is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is

Select Answer:

Visualized Solution

The Sigma Insight: Gaseous State

Solution Diagram

Analyzing the Setup

Imagine you have an open vessel sitting on a table. The temperature in the room is . Because the vessel is open, the gas inside is in direct contact with the atmosphere. This means the pressure inside the vessel will always be equal to the atmospheric pressure. So, pressure () is constant. The vessel itself is rigid, so its volume () is also constant.
Let's quickly convert our starting temperature to Kelvin by adding , giving us an initial temperature .

The Master Equation

Now, let's bring in our trusty tool, the ideal gas equation:
We just established that pressure and volume are constant. And of course, the universal gas constant is always constant. If we rearrange the equation to solve for , we get:
Since , , and are all constants, we can clearly see that the number of moles is inversely proportional to the temperature . This gives us a beautiful relationship:
This is the master key to solving our problem.

Tracking the Escaping Gas

The problem tells us that the vessel is heated until of the air escapes. Let's say initially, we had moles of air inside the vessel. That's our .
If of those moles escape into the atmosphere, how many moles are left inside? We have to be careful here; our gas law applies to the gas remaining inside the vessel. So, we subtract from the initial :
This remaining amount is our final number of moles, .

Final Calculation

We have all our pieces; now let's put them together. We substitute our values into the master equation:
Look at the equation now. We have the unknown variable on both sides. This is a classic physics problem trick! We don't even need to know the exact initial number of moles because will simply cancel out.
After canceling from both sides, we are left with:
To find , we need to isolate it. We multiply by and then divide by :
The vessel was heated to , which perfectly matches option (b).
Before we wrap up, think about the core concept here. In any open vessel problem, the pressure and volume are your constants, leading to moles being inversely proportional to temperature. If the question had asked for the answer in Celsius, we would just subtract from to get . Always pay attention to the units in the options!

Similar Questions

JEE Advanced 2018
LEVELJEE Advanced

A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does NOT allow the gas to leak across (Figure 2), the volume (in ) of the compartment A after the system attains equilibrium is_______.

JEE Main 2021
LEVELJEE Main

An LPG cylinder contains gas at a pressure of at . The cylinder can withstand the pressure of . The room in which the cylinder is kept catches fire. The minimum temperature at which the bursting of cylinder will take place is ……… . (Nearest integer)

JEE Main 2016
LEVELJEE Main

Two closed bulbs of equal volume () containing an ideal gas initially at pressure and temperature are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to . The final pressure is

(A)
(B)
(C)
(D)
LEVELJEE Main

Equal masses of methane and oxygen are mixed in an empty container at . The fraction of the total pressure exerted by oxygen is

(A)
(B)
(C)
(D)
JEE Main 2021
LEVELJEE Main

A car tyre is filled with nitrogen gas at at . It will burst if pressure exceeds . The temperature in at which the car tyre will burst is ......... (Rounded-off to the nearest integer).

JEE Main 2019
LEVELJEE Main

0.5 moles of gas A and moles of gas B exert a pressure of in a container of volume at . Given is the gas constant in , is

(A)
(B)
(C)
(D)
JEE Main 2021
LEVELJEE Main

A certain gas obeys . The value of is . The value of is ............ . ( compressibility factor)

JEE Advanced 2016
LEVELJEE Advanced

The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times. The value of x is

JEE Main 2021
LEVELJEE Main

An empty LPG cylinder weight . When full, it weight and shows a pressure of . In the course of use at ambient temperature, the mass of the cylinder is reduced to . The final pressure inside of the cylinder is ......... . (Nearest integer) (Assume LPG of be an ideal gas)

JEE Main 2021
LEVELBoard

Which one of the following is the correct vs plot at constant temperature for an ideal gas ? ( and stand for pressure and volume of the gas respectively)

(A)
(B)
(C)
(D)