The volume of gas A is twice that of gas B. The compressibility factor of gas A is thrice than that of gas B at the same temperature. The pressures of the gases for equal number of moles are :
a.) 2${P_A}$ = 3${P_B}$
b.) ${P_A}$ = 3${P_B}$
c.) ${P_A}$ = 2${P_B}$
d.) 3${P_A}$ = 2${P_B}$
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Hint: The compressibility factor is the correction factor which explains the behaviour of deviation of a real gas from ideal gas. Its value can be calculated as -
Z = $\dfrac{{PV}}{{nRT}}$
Where Z = compressibility factor
P = pressure of gas
V = volume of gas
‘n’ = number of moles of gas
R = gas constant
T = temperature of the gas
By filling the values, we can get the Z and comparing the values of Z, we can get our result.
Complete step by step answer:
Let us start by writing what is given to us and what we need to find.
Thus, Given :
Volume of gas A = 2 Volume of gas B
i.e. ${V_A}$ = 2${V_B}$
Compressibility factor of gas A = 3 Compressibility factor of gas B
This means, ${Z_A}$ = 3${Z_B}$
Temperature is same
So, ${T_A}$ = ${T_B}$= T
Equal number of moles
Thus, ${n_A}$ = ${n_B}$= n
To find :
Pressures of the gases
We have that the compressibility factor Z is equal to -
Z = $\dfrac{{PV}}{{nRT}}$
Where Z = compressibility factor
P = pressure of gas
V = volume of gas
‘n’ = number of moles of gas
R = gas constant
T = temperature of the gas
By, filling the values of all the factors in the formula for both the gases; we can find the value of pressure as -
${Z_A}$= $\dfrac{{{P_A}{V_A}}}{{{n_A}R{T_A}}}$
${Z_A}$ = $\dfrac{{{P_A}{V_A}}}{{nRT}}$
${Z_B}$ = $\dfrac{{{P_B}{V_B}}}{{{n_B}R{T_B}}}$
${Z_B}$ = $\dfrac{{{P_B}{V_B}}}{{nRT}}$
Further, we have
Compressibility factor of gas A = 3 Compressibility factor of gas B
Thus, ${Z_A}$ = 3${Z_B}$
$\dfrac{{{P_A}{V_A}}}{{nRT}}$= $\dfrac{{3{P_B}{V_B}}}{{nRT}}$
${P_A}{V_A}$= 3${P_B}{V_B}$
We have been given that ${V_A}$ = 2${V_B}$
So, filling value; we get
${P_A}$ 2${V_B}$= 3${P_B}{V_B}$
2${P_A}$ = 3${P_B}$
So, the option a.) is the correct answer.
Note: It must be noted that the R is gas constant and has value of 8.314 $J \cdot {K^{ - 1}}mo{l^{ - 1}}$. Its value remains constant for all gases at the same temperature. The ideal gas is one that follows the ideal gas equation. In reality, no gas follows the ideal gas equation as such. There are deviations from ideality. The z factor describes all these deviations of real gases from the ideal gas equation.
Z = $\dfrac{{PV}}{{nRT}}$
Where Z = compressibility factor
P = pressure of gas
V = volume of gas
‘n’ = number of moles of gas
R = gas constant
T = temperature of the gas
By filling the values, we can get the Z and comparing the values of Z, we can get our result.
Complete step by step answer:
Let us start by writing what is given to us and what we need to find.
Thus, Given :
Volume of gas A = 2 Volume of gas B
i.e. ${V_A}$ = 2${V_B}$
Compressibility factor of gas A = 3 Compressibility factor of gas B
This means, ${Z_A}$ = 3${Z_B}$
Temperature is same
So, ${T_A}$ = ${T_B}$= T
Equal number of moles
Thus, ${n_A}$ = ${n_B}$= n
To find :
Pressures of the gases
We have that the compressibility factor Z is equal to -
Z = $\dfrac{{PV}}{{nRT}}$
Where Z = compressibility factor
P = pressure of gas
V = volume of gas
‘n’ = number of moles of gas
R = gas constant
T = temperature of the gas
By, filling the values of all the factors in the formula for both the gases; we can find the value of pressure as -
${Z_A}$= $\dfrac{{{P_A}{V_A}}}{{{n_A}R{T_A}}}$
${Z_A}$ = $\dfrac{{{P_A}{V_A}}}{{nRT}}$
${Z_B}$ = $\dfrac{{{P_B}{V_B}}}{{{n_B}R{T_B}}}$
${Z_B}$ = $\dfrac{{{P_B}{V_B}}}{{nRT}}$
Further, we have
Compressibility factor of gas A = 3 Compressibility factor of gas B
Thus, ${Z_A}$ = 3${Z_B}$
$\dfrac{{{P_A}{V_A}}}{{nRT}}$= $\dfrac{{3{P_B}{V_B}}}{{nRT}}$
${P_A}{V_A}$= 3${P_B}{V_B}$
We have been given that ${V_A}$ = 2${V_B}$
So, filling value; we get
${P_A}$ 2${V_B}$= 3${P_B}{V_B}$
2${P_A}$ = 3${P_B}$
So, the option a.) is the correct answer.
Note: It must be noted that the R is gas constant and has value of 8.314 $J \cdot {K^{ - 1}}mo{l^{ - 1}}$. Its value remains constant for all gases at the same temperature. The ideal gas is one that follows the ideal gas equation. In reality, no gas follows the ideal gas equation as such. There are deviations from ideality. The z factor describes all these deviations of real gases from the ideal gas equation.