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What is the equivalent value of $\dfrac{PV}{T}$ for one mole of an ideal gas?
A. $2{\text{ Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$
B. ${\text{8}}{\text{.3 Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$
C. $4.2{\text{ Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$
D. $2{\text{ calmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$


Answer
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Hint:According to ideal gas law, $PV = nRT$ , where $P$ is the pressure of the gas, $V$ is its volume, $n$ is the number of moles, $T$ is temperature and $R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ . Use this law to get the required answer.



Formula used:
 Ideal gas law,
$PV = nRT$ … (1)
Here $P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles,
$T$ is the temperature and
$R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ .

Complete answer:
According to ideal gas law,
$PV = nRT$ … (1)
In the above question, it is given that there is one mole of gas.
Thus, $n = 1$ .
Now, by using ideal gas law, substituting the values in formula (1) we get:
$PV = RT$
Rearranging,
$\dfrac{PV}{T}$ = R = 8.314${\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$
In the given options, the value closest to this is ${\text{8}}{\text{.3 Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$
Thus, the correct option is B.



Note: Ideal gas law, also known as perfect gas law, states that the product of pressure and volume of one mole of an ideal gas is equivalent to the product of $R$ , the ideal gas constant, and the absolute temperature. This law has its own limitations.