
A gas at NTP is suddenly compressed to one-fourth of its original volume. If $\lambda $ is supposed to be $\dfrac{3}{2}$, then the final pressure is
(A) 4 atmosphere
(B) $\dfrac{3}{2}$ atmosphere
(C) 8 atmosphere
(D) $\dfrac{1}{4}$ atmosphere
Answer
535.2k+ views
Hint: When a gas is suddenly compressed, it does not have enough time to release heat to its surroundings. When there is no heat transfer to or from a system, the process is known as an adiabatic process.
Complete step-by-step answer:
Now since we know that the sudden compression of gas is an adiabatic process, the relation between pressure and volume can be expressed by
\[{{P}_{0}}V_{0}^{\gamma }=P{{V}^{\gamma }}=\text{constant}\]
Here, ${{P}_{0}}$= pressure of gas before compression
${{V}_{0}}$= volume of gas before compression
P= pressure of gas after compression
V= volume of gas after compression
$\gamma $= adiabatic index of the gas
It is given to us that $\gamma =\dfrac{3}{2}$, $V=\dfrac{1}{4}{{V}_{0}}$, and since the gas is at NTP ${{P}_{0}}=1atm$. So, by substituting these values in the expression, we can calculate the final pressure of the gas.
\[
\Rightarrow P={{P}_{0}}{{\left( \dfrac{{{V}_{0}}}{V} \right)}^{\gamma }} \\
\Rightarrow P=1\times {{\left( \dfrac{{{V}_{0}}}{\dfrac{1}{4}{{V}_{0}}} \right)}^{\dfrac{3}{2}}} \\
\Rightarrow P={{4}^{\dfrac{3}{2}}} \\
\Rightarrow P=8atm
\]
So, when a gas at NTP is suddenly compressed to one-fourth of its volume and $\gamma =\dfrac{3}{2}$, the final temperature of the gas will be equal to option (C) 8 atmosphere.
Note: It is easy to confuse between the isothermal and the adiabatic processes.
The main difference between the two is that while in the adiabatic process there is no transfer of heat to or from the surroundings, to make the overall temperature constant, it is accompanied by a change in the temperature of the system.
On the other hand, in an isothermal process, while there is no change in temperature of the system, to make the overall temperature constant, there is a transfer of heat to and from the surrounding.
Complete step-by-step answer:
Now since we know that the sudden compression of gas is an adiabatic process, the relation between pressure and volume can be expressed by
\[{{P}_{0}}V_{0}^{\gamma }=P{{V}^{\gamma }}=\text{constant}\]
Here, ${{P}_{0}}$= pressure of gas before compression
${{V}_{0}}$= volume of gas before compression
P= pressure of gas after compression
V= volume of gas after compression
$\gamma $= adiabatic index of the gas
It is given to us that $\gamma =\dfrac{3}{2}$, $V=\dfrac{1}{4}{{V}_{0}}$, and since the gas is at NTP ${{P}_{0}}=1atm$. So, by substituting these values in the expression, we can calculate the final pressure of the gas.
\[
\Rightarrow P={{P}_{0}}{{\left( \dfrac{{{V}_{0}}}{V} \right)}^{\gamma }} \\
\Rightarrow P=1\times {{\left( \dfrac{{{V}_{0}}}{\dfrac{1}{4}{{V}_{0}}} \right)}^{\dfrac{3}{2}}} \\
\Rightarrow P={{4}^{\dfrac{3}{2}}} \\
\Rightarrow P=8atm
\]
So, when a gas at NTP is suddenly compressed to one-fourth of its volume and $\gamma =\dfrac{3}{2}$, the final temperature of the gas will be equal to option (C) 8 atmosphere.
Note: It is easy to confuse between the isothermal and the adiabatic processes.
The main difference between the two is that while in the adiabatic process there is no transfer of heat to or from the surroundings, to make the overall temperature constant, it is accompanied by a change in the temperature of the system.
On the other hand, in an isothermal process, while there is no change in temperature of the system, to make the overall temperature constant, there is a transfer of heat to and from the surrounding.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

