1/2 mole of helium is contained in a container at STP. How much heat energy is needed to double the pressure of the gas (volume is constant) heat capacity of has is 3J/g/K
(A) 1436
(B) 736
(C) 1638
(D) 5698
Answer
300.3k+ views
Hint: The pressure of a given mass of gas is directly proportional to the absolute temperature provided that the volume is kept constant. Use that relation to calculate the temperature difference of the two different states of the system (the high pressure state and the stp pressure state). Standard temperature is considered as zero degree Celsius or 273 Kelvin.
Formula used: In this solution we will be using the following formulae;
\[P = kT\] where \[P\] is the pressure of an ideal gas at a certain state, \[T\] and is the absolute temperature of the gas at the same state, \[k\] is a proportionality constant.
\[Q = nM{c_v}\Delta T\] where \[Q\] is the heat absorbed by a gas,\[M\] is the molar mass,\[n\] is the number of moles of the gas, \[{c_v}\] is the specific heat capacity at constant volume and \[\Delta T\] is the difference in temperature after a particular amount of absorbed heat.
Complete Step-by-Step solution:
To calculate for heat needed, we must first calculate the temperature necessary for the condition to hold.
Generally, it is known that the pressure of a gas is directly proportional to temperature of the gas provided the volume is held constant. Hence
\[P = kT\] where \[P\] is the pressure of an ideal gas at a certain state, \[T\] and is the absolute temperature of the gas at the same state, \[k\] is a proportionality constant.
Hence, for the first state, we have
\[{P_1} = k{T_1} = 273k\] (since the standard temperature is 373 K)
\[ \Rightarrow k = \dfrac{{{P_1}}}{{273}}\]
For the second state, we have
\[ \Rightarrow 2{P_1} = k{T_2}\]
Inserting the known \[k\] expression, we have
\[2{P_1} = \left( {\dfrac{{{P_1}}}{{273}}} \right){T_2}\]
\[ \Rightarrow 2 = \dfrac{{{T_2}}}{{273}}\]
Hence, we have
\[{T_2} = 2 \times 273 = 546K\]
Now, the heat required can be given by
\[Q = nM{c_v}\Delta T\] where \[n\] is the number of moles of the gas,\[M\] is the molar mass,\[{c_v}\] is the specific heat capacity at constant volume and \[\Delta T\] is the difference in temperature after a particular amount of absorbed heat.
Hence, we have
\[Q = \left( {0.5} \right)4\left( 3 \right)\left( {546 - 273} \right)\]
By computation,
\[Q = 1638J\]
The correct option is C
Note: Alternatively, instead for calculating for \[k\] and then using its value in the second state, we could simply divide state 1 by state 2, and hence, we have
\[\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{T_1}}}{{{T_2}}}\]
Then by making \[{T_2}\] subject of formula, we have
\[{T_2} = \dfrac{{{P_2}{T_1}}}{{{P_1}}}\]
Hence, by inserting the known values and expressions, we have
\[{T_2} = \dfrac{{2{P_1}\left( {273} \right)}}{{{P_1}}} = 546K\]
Which is identical to what is gotten above.
Formula used: In this solution we will be using the following formulae;
\[P = kT\] where \[P\] is the pressure of an ideal gas at a certain state, \[T\] and is the absolute temperature of the gas at the same state, \[k\] is a proportionality constant.
\[Q = nM{c_v}\Delta T\] where \[Q\] is the heat absorbed by a gas,\[M\] is the molar mass,\[n\] is the number of moles of the gas, \[{c_v}\] is the specific heat capacity at constant volume and \[\Delta T\] is the difference in temperature after a particular amount of absorbed heat.
Complete Step-by-Step solution:
To calculate for heat needed, we must first calculate the temperature necessary for the condition to hold.
Generally, it is known that the pressure of a gas is directly proportional to temperature of the gas provided the volume is held constant. Hence
\[P = kT\] where \[P\] is the pressure of an ideal gas at a certain state, \[T\] and is the absolute temperature of the gas at the same state, \[k\] is a proportionality constant.
Hence, for the first state, we have
\[{P_1} = k{T_1} = 273k\] (since the standard temperature is 373 K)
\[ \Rightarrow k = \dfrac{{{P_1}}}{{273}}\]
For the second state, we have
\[ \Rightarrow 2{P_1} = k{T_2}\]
Inserting the known \[k\] expression, we have
\[2{P_1} = \left( {\dfrac{{{P_1}}}{{273}}} \right){T_2}\]
\[ \Rightarrow 2 = \dfrac{{{T_2}}}{{273}}\]
Hence, we have
\[{T_2} = 2 \times 273 = 546K\]
Now, the heat required can be given by
\[Q = nM{c_v}\Delta T\] where \[n\] is the number of moles of the gas,\[M\] is the molar mass,\[{c_v}\] is the specific heat capacity at constant volume and \[\Delta T\] is the difference in temperature after a particular amount of absorbed heat.
Hence, we have
\[Q = \left( {0.5} \right)4\left( 3 \right)\left( {546 - 273} \right)\]
By computation,
\[Q = 1638J\]
The correct option is C
Note: Alternatively, instead for calculating for \[k\] and then using its value in the second state, we could simply divide state 1 by state 2, and hence, we have
\[\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{T_1}}}{{{T_2}}}\]
Then by making \[{T_2}\] subject of formula, we have
\[{T_2} = \dfrac{{{P_2}{T_1}}}{{{P_1}}}\]
Hence, by inserting the known values and expressions, we have
\[{T_2} = \dfrac{{2{P_1}\left( {273} \right)}}{{{P_1}}} = 546K\]
Which is identical to what is gotten above.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics 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
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2026-27 Free PDF Download (Sign-in Required)

