When air at $300\,K$ and $1\,atm$ pressure is compressed to one-twentieth of its original volume and $30\,atm$ pressure, the new temperature will be?
Answer
549.6k+ views
Hint:In this question, we will first list out the values given to us and then apply the appropriate values in the gas equation given as $\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$ where the subscripts represent two different conditions. Here $P$ is the pressure, $V$ is the volume and $T$ is the temperature of the gas. The number of moles of gas will remain constant.
Complete step by step answer:
Let the initial pressure be ${P_1}$ , initial volume be ${V_1}$ and the initial temperature be ${T_1}$. In the question it is given that ${P_1} = 1\,atm$ and ${T_1} = 300\,K$. Now the final pressure is given as ${P_2}$ , final volume by ${V_2}$ and the final temperature by ${T_2}$. In the question it is given that ${P_2} = 30\,atm$ and ${V_2} = \dfrac{{{V_1}}}{{20}}$. Now that we have listed the variables and their values as given in the question, we shall move further on writing the appropriate equation in which we can put all the values known to us and get the answer. The gas equation says that for a given number of moles of a particular gas, the quantity $\dfrac{{PV}}{T}$ remains constant. Here P is the pressure, V is the volume and T is the temperature of the gas.
This means that $\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$
Substituting the known values we have,
$\dfrac{{1 \times {V_1}}}{{300}} = \dfrac{{30 \times \dfrac{{{V_1}}}{{20}}}}{{{T_2}}}$
Cancelling out the common terms we have,
$\dfrac{1}{{300}} = \dfrac{{30}}{{20{T_2}}}$
This can be rewritten as
${T_2} = \dfrac{{30}}{{20}} \times 300$
Further solving this we get,
$\therefore {T_2} = 450\,K$
Hence, the new temperature will be ${T_2} = 450\,K$.
Note:The gas equation is based on the assumptions that there are no intermolecular forces between the particles of a gas. Combining the laws of Charles, Boyle and Gay-Lussac gives the combined gas law or the gas equation. Here $T$ is the absolute temperature of the gas and must be always taken in Kelvin. Other quantities can be used in different units they have.
Complete step by step answer:
Let the initial pressure be ${P_1}$ , initial volume be ${V_1}$ and the initial temperature be ${T_1}$. In the question it is given that ${P_1} = 1\,atm$ and ${T_1} = 300\,K$. Now the final pressure is given as ${P_2}$ , final volume by ${V_2}$ and the final temperature by ${T_2}$. In the question it is given that ${P_2} = 30\,atm$ and ${V_2} = \dfrac{{{V_1}}}{{20}}$. Now that we have listed the variables and their values as given in the question, we shall move further on writing the appropriate equation in which we can put all the values known to us and get the answer. The gas equation says that for a given number of moles of a particular gas, the quantity $\dfrac{{PV}}{T}$ remains constant. Here P is the pressure, V is the volume and T is the temperature of the gas.
This means that $\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$
Substituting the known values we have,
$\dfrac{{1 \times {V_1}}}{{300}} = \dfrac{{30 \times \dfrac{{{V_1}}}{{20}}}}{{{T_2}}}$
Cancelling out the common terms we have,
$\dfrac{1}{{300}} = \dfrac{{30}}{{20{T_2}}}$
This can be rewritten as
${T_2} = \dfrac{{30}}{{20}} \times 300$
Further solving this we get,
$\therefore {T_2} = 450\,K$
Hence, the new temperature will be ${T_2} = 450\,K$.
Note:The gas equation is based on the assumptions that there are no intermolecular forces between the particles of a gas. Combining the laws of Charles, Boyle and Gay-Lussac gives the combined gas law or the gas equation. Here $T$ is the absolute temperature of the gas and must be always taken in Kelvin. Other quantities can be used in different units they have.
Recently Updated Pages
Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Father s age is three times the sum of the ages of-class-11-maths-CBSE

Give a comparative account of the classes of kingdom class 11 biology CBSE

The ceiling of a long hall is 25m high What is the class 11 physics CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Name the Largest and the Smallest Cell in the Human Body ?

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

Find the value of the expression given below sin 30circ class 11 maths CBSE

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

Two of the body parts which do not appear in MRI are class 11 biology CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

