How much volume does one mole of gas occupy at NTP?
Answer
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Hint: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, $PV = nRT$ where $P$ is the pressure of the ideal gas, $V$ is the volume of the given sample, $n$ is the number of moles present in the given sample, $T$ is the temperature and $R$ 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 $T = 20^\circ {\text{C}}$ .
The pressure at NTP is taken to be $P = 1{\text{atm}}$ .
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 $PV = nRT$ ------------- (1)
where $P$ is the pressure of the ideal gas, $V$ is the volume of the given sample, $n$ is the number of moles present in the given sample, $T$ is the temperature and $R$ is the gas constant.
Rewriting equation (1) in terms of the volume we get, $V = \dfrac{{nRT}}{P}$ ----------- (2)
Substituting for $n = 1$ , $T = 293{\text{K}}$ , $P = 101325{\text{Pa}}$ and $R = 8 \cdot 314{\text{Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ in equation (2) we get, $V = \dfrac{{1 \times 8 \cdot 314 \times 293}}{{101325}} = 24 \cdot 12 \times {10^{ - 3}}{{\text{m}}^3}$
Thus the volume of one mole of gas at NTP is obtained as $V = 24 \cdot 04 \times {10^{ - 3}}{{\text{m}}^3} = 24 \cdot 04{\text{L}}$ .
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 $T = 20^\circ {\text{C}} + 273{\text{K}} = 293{\text{K}}$ and $P = 1{\text{atm}} = 101325{\text{Pa}}$ before substituting in equation (2). STP stands for standard temperature and pressure. The pressure at STP is the same $P = 1{\text{atm}}$ but the temperature at STP will be $T = 0^\circ {\text{C}} = 273{\text{K}}$ .
Formula used:
-The ideal gas equation is given by, $PV = nRT$ where $P$ is the pressure of the ideal gas, $V$ is the volume of the given sample, $n$ is the number of moles present in the given sample, $T$ is the temperature and $R$ 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 $T = 20^\circ {\text{C}}$ .
The pressure at NTP is taken to be $P = 1{\text{atm}}$ .
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 $PV = nRT$ ------------- (1)
where $P$ is the pressure of the ideal gas, $V$ is the volume of the given sample, $n$ is the number of moles present in the given sample, $T$ is the temperature and $R$ is the gas constant.
Rewriting equation (1) in terms of the volume we get, $V = \dfrac{{nRT}}{P}$ ----------- (2)
Substituting for $n = 1$ , $T = 293{\text{K}}$ , $P = 101325{\text{Pa}}$ and $R = 8 \cdot 314{\text{Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ in equation (2) we get, $V = \dfrac{{1 \times 8 \cdot 314 \times 293}}{{101325}} = 24 \cdot 12 \times {10^{ - 3}}{{\text{m}}^3}$
Thus the volume of one mole of gas at NTP is obtained as $V = 24 \cdot 04 \times {10^{ - 3}}{{\text{m}}^3} = 24 \cdot 04{\text{L}}$ .
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 $T = 20^\circ {\text{C}} + 273{\text{K}} = 293{\text{K}}$ and $P = 1{\text{atm}} = 101325{\text{Pa}}$ before substituting in equation (2). STP stands for standard temperature and pressure. The pressure at STP is the same $P = 1{\text{atm}}$ but the temperature at STP will be $T = 0^\circ {\text{C}} = 273{\text{K}}$ .
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