Question

What if STP was defined at normal room temperature ($27{}^\circ C$) and 1 atm. How would this affect molar volume of an ideal gas.
(A) Same
(B) Increase
(C) Decrease
(D) Cannot be predicted

Answer 1

The STP (Standard Temperature and Pressure) as the name suggests it is defined at standard temperature and pressure i.e. 0${}^\circ C$ and 1 atm.
Now, if STP is defined at some other temperature or pressure, molar volume will change according to the below given equations.

Complete step by step answer:
Let us first understand the concepts of STP and the relationship of STP with molar volume of gas.
As we know, STP is defined for the gases hence, the molar volume calculations will be normally based on the gas laws. Thus,
At STP, the molar volume is the volume occupied by one mole of a chemical element or compound. This can be calculated as the ratio of molar mass and mass density.
${{V}_{m}}=\dfrac{M}{\rho }$
where,
${{V}_{m}}$ = molar volume
M = molar mass
$\rho$ = mass density
The molar volume of gas is the volume of one mole of any gas at specific temperature and pressure. The standard temperature used for calculation is 273 K or 0${}^\circ C$ and standard pressure is 1 atmosphere or 760 mm Hg.

-For ideal gases,
$PV=nRT$
At molar volume,

& P{{V}_{m}}=nRT \\\ & {{V}_{m}}=\dfrac{nRT}{P} \\\ & {{V}_{m}}=\dfrac{1mol\times 0.08205Latm/\left( mol.K \right)\times 273K}{1atm} \\\ & \therefore {{V}_{m}}=22.39L\approx 22.4L \\\ \end{aligned}$$ - For real gases, The molar volume depends on the constants ‘a’ and ‘b’ as, $$\left[ P+\dfrac{a{{n}^{2}}}{{{v}^{2}}} \right]\left( V-nb \right)=nRT$$ Thus, Experimentally, one mole of gas occupies a volume of 22.4 L at STP. Now, when molar volume is calculated at room temperature i.e. 27${}^\circ C$(given). ${{V}_{m}}$ at room temperature is given as, $${{V}_{m}}=\dfrac{1mol\times 0.08205 L atm/\left( mol.K \right)\times \left( 273+27 \right)K}{1atm}$$= 24.615 L Thus, the molar volume will increase if STP is considered at room temperature i.e. 27${}^\circ C$ as given. **So, the correct answer is “Option B”.** **Note:** Do note that when the standard temperature will change, there would be changes in multiplying factors for a given ${{V}_{m}}$ formula. So, there will always be a change in molar volume at considered standard temperatures or pressures than the actual parameters.