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Question: How much volume does one mole of gas occupy at NTP?...

How much volume does one mole of gas occupy at NTP?

Explanation

Solution

NTP stands for normal temperature and pressure. The ideal gas equation which is a relation connecting the pressure, volume, temperature and number of moles present in a sample of an ideal gas can be employed to determine the volume of one mole of gas.

Formula used:
-The ideal gas equation is given by, PV=nRTPV = nRT where PP is the pressure of the ideal gas, VV is the volume of the given sample, nn is the number of moles present in the given sample, TT is the temperature and RR is the gas constant.

Complete step by step answer.
Step 1: Mention the temperature and pressure of a gas at NTP.
The temperature at NTP is taken to be T=20CT = 20^\circ {\text{C}} .
The pressure at NTP is taken to be P=1atmP = 1{\text{atm}} .
We have to determine the volume of one mole of gas.
Step 2: Based on the ideal gas equation, obtain the volume of one mole of an ideal gas at NTP.
The ideal gas equation can be expressed as PV=nRTPV = nRT ------------- (1)
where PP is the pressure of the ideal gas, VV is the volume of the given sample, nn is the number of moles present in the given sample, TT is the temperature and RR is the gas constant.
Rewriting equation (1) in terms of the volume we get, V=nRTPV = \dfrac{{nRT}}{P} ----------- (2)
Substituting for n=1n = 1 , T=293KT = 293{\text{K}} , P=101325PaP = 101325{\text{Pa}} and R=8314Jmol1K1R = 8 \cdot 314{\text{Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}} in equation (2) we get, V=1×8314×293101325=2412×103m3V = \dfrac{{1 \times 8 \cdot 314 \times 293}}{{101325}} = 24 \cdot 12 \times {10^{ - 3}}{{\text{m}}^3}
Thus the volume of one mole of gas at NTP is obtained as V=2404×103m3=2404LV = 24 \cdot 04 \times {10^{ - 3}}{{\text{m}}^3} = 24 \cdot 04{\text{L}} .

Note: Here we obtained the volume of one mole of an ideal gas. For a real gas, the value will be slightly different. The S.I unit of volume is cubic-meter and so we converted the temperature and pressure into their respective S.I units as T=20C+273K=293KT = 20^\circ {\text{C}} + 273{\text{K}} = 293{\text{K}} and P=1atm=101325PaP = 1{\text{atm}} = 101325{\text{Pa}} before substituting in equation (2). STP stands for standard temperature and pressure. The pressure at STP is the same P=1atmP = 1{\text{atm}} but the temperature at STP will be T=0C=273KT = 0^\circ {\text{C}} = 273{\text{K}} .