An open vessel at 27 degree Celsius is heated until two fifths 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:
a. 720\[{}^\circ C\]
b. 500\[{}^\circ C\]
c. 750\[{}^\circ C\]
d. 550 K
Answer
591.9k+ views
Hint: To solve this question, look at the parameters given to us in the question. Since, pressure (P), volume (V) and real gas constant (R) remain constant, relate the number of moles and temperature by using the ideal gas equation.
Complete step by step answer:
According to the question, there is an open vessel with a temperature equal to 27 deg Celsius.
27\[{}^\circ C\] = 273.15 + 27 = 300K.
Also, two-fifth of air escapes. Therefore, we can say that there is a change in moles of the gas.
So, let the initial moles in air be ‘n1’ and the number of moles after two fifths of gas escaped be ‘n2’.
Let n1 = 1 mole
So, n2 = n1 – (2/5) = 1 – (2/5) = 3/5 moles.
According to the question we can say that the volume remains constant.
Ideal gas equation relates PV = nRT.
Since, pressure (P), volume (V) and real gas constant (R) are constant, we can relate the number of moles and temperature as –
nT = constant
\[{{n}_{1}}{{T}_{1}}={{n}_{2}}{{T}_{2}}\]= constant
Now, putting the values of moles and temperature we get –
\[\begin{align}
& (1)(300)=\left( \dfrac{3}{5} \right){{T}_{2}} \\
& {{T}_{2}}=\dfrac{\text{300x}5}{3}K \\
& {{T}_{2}}=500K \\
\end{align}\]
Therefore, the answer is – option (d). The temperature at which the vessel has been heated is 500K.
Additional Information:
1 mole of any gas at STP occupies a volume of 22.4 L.
Note: Ideal gas equation is the equation of state of an ideal gas (hypothetical). It is an approximation of the behaviour of gases under ideal conditions.It is a combination of empirical laws like Boyle’s law, Charles law, Gay-Lussac’s law and Avogadro’s law.
Complete step by step answer:
According to the question, there is an open vessel with a temperature equal to 27 deg Celsius.
27\[{}^\circ C\] = 273.15 + 27 = 300K.
Also, two-fifth of air escapes. Therefore, we can say that there is a change in moles of the gas.
So, let the initial moles in air be ‘n1’ and the number of moles after two fifths of gas escaped be ‘n2’.
Let n1 = 1 mole
So, n2 = n1 – (2/5) = 1 – (2/5) = 3/5 moles.
According to the question we can say that the volume remains constant.
Ideal gas equation relates PV = nRT.
Since, pressure (P), volume (V) and real gas constant (R) are constant, we can relate the number of moles and temperature as –
nT = constant
\[{{n}_{1}}{{T}_{1}}={{n}_{2}}{{T}_{2}}\]= constant
Now, putting the values of moles and temperature we get –
\[\begin{align}
& (1)(300)=\left( \dfrac{3}{5} \right){{T}_{2}} \\
& {{T}_{2}}=\dfrac{\text{300x}5}{3}K \\
& {{T}_{2}}=500K \\
\end{align}\]
Therefore, the answer is – option (d). The temperature at which the vessel has been heated is 500K.
Additional Information:
1 mole of any gas at STP occupies a volume of 22.4 L.
Note: Ideal gas equation is the equation of state of an ideal gas (hypothetical). It is an approximation of the behaviour of gases under ideal conditions.It is a combination of empirical laws like Boyle’s law, Charles law, Gay-Lussac’s law and Avogadro’s law.
Recently Updated Pages
Normality of 03 M phosphorus acid H3PO3 is A 05 B 06 class 11 chemistry JEE_Main

A molecule with highest bond energy A Fluorine B Chlorine class 11 chemistry JEE_Main

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main

Covalent compounds generally have low melting and boiling class 11 chemistry JEE_Main

When an acid reacts with a metal carbonate or metal class 11 chemistry JEE_Main

The degeneracy of hydrogen atom that has equal energy class 11 chemistry JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

NCERT Solutions For Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts Of Chemistry - 2026-27 Free PDF Download (Sign-in Required)

Effective Nuclear Charge for JEE

Hybridisation in Chemistry – Concept, Types & Applications

CBSE Notes Class 11 Chemistry Chapter 5 - Thermodynamics - 2026-27 PDF Download (Login Required)

Isoelectronic Species: Definition, Examples & Importance

