
The isothermal bulk modulus of a perfect gas at normal pressure is
A. $1.013 \times {10^{5\,}}\,N/{m^2}$
B. $1.013 \times {10^{6\,}}\,N/{m^2}$
C. $1.013 \times {10^{ - 11\,}}\,N/{m^2}$
D. $1.013 \times {10^{11\,}}\,N/{m^2}$
Answer
162.6k+ views
Hint:
In a thermodynamic system, the ratio of the substance's volume to an infinite pressure rise is known as the bulk modulus. Compressibility is the reciprocal of the bulk modulus. Therefore, the isothermal bulk modulus for a perfect gas can be determined by differentiating an ideal gas equation $PV = nRT$ at a constant temperature.
Formula used:
An Ideal-Gas Equation, $PV = nRT$
where $P$ stands for pressure, $V$ for volume, $n$ for moles, $R$ for the universal gas constant, and $T$ for temperature.
and, the product rule of differentiation $\dfrac{d}{{dx}}\left( {I \cdot II} \right) = I \cdot \dfrac{d}{{dx}}\left( {II} \right) + II \cdot \dfrac{d}{{dx}}\left( I \right)$
Complete step by step solution:
In Isothermal process, $\,Change{\text{ }}in{\text{ }}Temperature = \Delta T = 0$
i.e., $T = constant$
Now, the expression for an ideal-gas equation can be stated as: -
$PV = nRT$
In an Isothermal expansion of a perfect gas, $T = constant$
$ \Rightarrow PV = nR(constant)$
As, $n\,and\,R$ are constant for given ideal gas.
$ \Rightarrow PV = constant = k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)$
On differentiating and using product rule $\dfrac{d}{{dx}}\left( {I \cdot II} \right) = I \cdot \dfrac{d}{{dx}}\left( {II} \right) + II \cdot \dfrac{d}{{dx}}\left( I \right)$ in equation $(1)$, we get
$ \Rightarrow PdV + VdP = 0$ $\left\{ {\therefore \dfrac{d}{{dx}}(constant) = 0} \right\}$
$ \Rightarrow \dfrac{{dP}}{P} = - \dfrac{{dV}}{V}\,\left( {for{\text{ }}small{\text{ }}change} \right)$
For a large change,
$ \Rightarrow \dfrac{{\Delta P}}{P} = - \dfrac{{\Delta V}}{V}\,$
$ \Rightarrow P = \dfrac{{\Delta P}}{{\left( { - \dfrac{{\Delta V}}{V}\,} \right)}} = - \dfrac{{\Delta P}}{{\left( {\dfrac{{\Delta V}}{V}\,} \right)}} = {B_{isothermal}}$
Now, we know that the normal pressure for an ideal gas is $1.013 \times {10^{5\,}}\,N/{m^2}$. Therefore, the isothermal bulk modulus will be: -
$ \Rightarrow {B_{isothermal}} = P = 1.013 \times {10^{5\,}}\,N/{m^2}$
Thus, the isothermal bulk modulus of a perfect gas at normal pressure is $1.013 \times {10^{5\,}}\,N/{m^2}$.
Hence, the correct option is (A) $1.013 \times {10^{5\,}}\,N/{m^2}$ .
Note:
The ratio of the change in pressure to the fractional change in volume at constant temperature is known as the isothermal bulk modulus. In this problem, to determine the isothermal bulk modulus, the derivation of an ideal gas $PV = nRT$ is evaluated at a constant temperature.
In a thermodynamic system, the ratio of the substance's volume to an infinite pressure rise is known as the bulk modulus. Compressibility is the reciprocal of the bulk modulus. Therefore, the isothermal bulk modulus for a perfect gas can be determined by differentiating an ideal gas equation $PV = nRT$ at a constant temperature.
Formula used:
An Ideal-Gas Equation, $PV = nRT$
where $P$ stands for pressure, $V$ for volume, $n$ for moles, $R$ for the universal gas constant, and $T$ for temperature.
and, the product rule of differentiation $\dfrac{d}{{dx}}\left( {I \cdot II} \right) = I \cdot \dfrac{d}{{dx}}\left( {II} \right) + II \cdot \dfrac{d}{{dx}}\left( I \right)$
Complete step by step solution:
In Isothermal process, $\,Change{\text{ }}in{\text{ }}Temperature = \Delta T = 0$
i.e., $T = constant$
Now, the expression for an ideal-gas equation can be stated as: -
$PV = nRT$
In an Isothermal expansion of a perfect gas, $T = constant$
$ \Rightarrow PV = nR(constant)$
As, $n\,and\,R$ are constant for given ideal gas.
$ \Rightarrow PV = constant = k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)$
On differentiating and using product rule $\dfrac{d}{{dx}}\left( {I \cdot II} \right) = I \cdot \dfrac{d}{{dx}}\left( {II} \right) + II \cdot \dfrac{d}{{dx}}\left( I \right)$ in equation $(1)$, we get
$ \Rightarrow PdV + VdP = 0$ $\left\{ {\therefore \dfrac{d}{{dx}}(constant) = 0} \right\}$
$ \Rightarrow \dfrac{{dP}}{P} = - \dfrac{{dV}}{V}\,\left( {for{\text{ }}small{\text{ }}change} \right)$
For a large change,
$ \Rightarrow \dfrac{{\Delta P}}{P} = - \dfrac{{\Delta V}}{V}\,$
$ \Rightarrow P = \dfrac{{\Delta P}}{{\left( { - \dfrac{{\Delta V}}{V}\,} \right)}} = - \dfrac{{\Delta P}}{{\left( {\dfrac{{\Delta V}}{V}\,} \right)}} = {B_{isothermal}}$
Now, we know that the normal pressure for an ideal gas is $1.013 \times {10^{5\,}}\,N/{m^2}$. Therefore, the isothermal bulk modulus will be: -
$ \Rightarrow {B_{isothermal}} = P = 1.013 \times {10^{5\,}}\,N/{m^2}$
Thus, the isothermal bulk modulus of a perfect gas at normal pressure is $1.013 \times {10^{5\,}}\,N/{m^2}$.
Hence, the correct option is (A) $1.013 \times {10^{5\,}}\,N/{m^2}$ .
Note:
The ratio of the change in pressure to the fractional change in volume at constant temperature is known as the isothermal bulk modulus. In this problem, to determine the isothermal bulk modulus, the derivation of an ideal gas $PV = nRT$ is evaluated at a constant temperature.
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