Question

The value of ${\text{PV}}$ for 5.6 ${\text{L}}$ of an ideal gas 0.5 ${\text{RT}}$ at STP .If true enter 1 else 0.

Answer 1

An ideal gas is defined as the one in which all the collisions between the atoms or molecules are perfectly elastic and there are no intermolecular attractive forces. That is we can imagine ideal gas as a collection of perfectly hard spheres which collide but do not interact with each other. An ideal gas can be characterised by three state variables , absolute pressure ( ${\text{P}}$ ) , volume ( ${\text{V}}$ ) and absolute temperature ( ${\text{T}}$ ).
Formula used:
There are three primary gas laws , which deals with how gases behave with respect to pressure $\left( P \right)$, volume $(V)$ , temperature $(T)$ and amount. The three important gas law are given below
Boyle's law: ${\text{V}}\, \propto \,\dfrac{1}{{\text{P}}}\,$ , at constant ${\text{n}}\,$and ${\text{T}}$.
Charles law: ${\text{V}}\,\, \propto \,{\text{T}}$, at constant ${\text{n}}$ and ${\text{p}}$
Avogadro's law: ${\text{V}}\, \propto \,{\text{n}}$, at constant ${\text{P}}$ and ${\text{T}}$
Combining above gas laws , we get ${\text{V }}\, \propto \,\,\dfrac{{{\text{nT}}}}{{\text{P}}}$
By rearranging above formula we get ${\text{PV}}\, \propto \,{\text{nT}}$
Hence we can write it as ${\text{PV}}\,{\text{ = }}\,{\text{nRT}}$ , where ${\text{R}}$ is the gas constant.

Complete answer:
STP is used as a standard reference point for expression of the properties of an ideal gas , Where standard temperature is the freezing point of water and the standard pressure is one standard atmosphere. Which can be quantified as follows
Standard temperature: ${{\text{0}}^{\text{0}}}\,{\text{C}}\,{\text{ = }}\,{\text{ 273}}{\text{.15}}\,{\text{K}}$
Standard pressure: 1 atmosphere = ${\text{760 mm Hg = }}\,{\text{101}}{\text{.3 kPa}}$
Standard volume of 1 mole of an ideal gas at STP : ${\text{22}}{\text{.4}}\,{\text{L}}$
Ideal gas equation is given by ${\text{PV = nRT}}$
We have to calculate ${\text{PV}}$ for $5.6\;{\text{L}}$of an ideal gas
At STP , Standard volume of 1 mole of an ideal gas contains ${\text{22}}{\text{.4 L}}$.
$\therefore$No.of moles in $5.6\;{\text{L}}$ of ideal gas, ${\text{n}}$ = $\dfrac{{{\text{5}}{\text{.6L}}}}{{{\text{22}}{\text{.4L}}}}$
Applying the value of ${\text{n}}$ in ideal gas equation , we get
${\text{PV}}\,{\text{ = }}\dfrac{{5.6}}{{22.4}}{\text{RT}}$
${\text{PV}}\,{\text{ = 0}}{\text{.25 RT}}$
So here the value of ${\text{PV}}$ for an ${\text{5}}{\text{.6 L}}$ of an ideal gas is $0.25\,{\text{RT}}$ at STP.

The given statement in the question “ The value of ${\text{PV}}$ for ${\text{5}}{\text{.6 L}}$ of an ideal gas IS ${\text{0}}{\text{.5 RT}}$at STP “ is not true hence you can enter 0.

Note:
For an ideal gas all the internal energy is in the form of kinetic energy and any change in internal energy is accompanied by the change in temperature. An ideal gas is an idealized model for real gases that have sufficiently low densities. The condition for low density means that the molecules of the gas are so far apart that they do not interact.