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Question: There are \[4 \times {10^{24}}\] gas molecules in a vessel at 50 K temperature. The pressure of the ...

There are 4×10244 \times {10^{24}} gas molecules in a vessel at 50 K temperature. The pressure of the gas in the vessel is 0.03 atmospheric. Calculate the volume of the vessel.

Explanation

Solution

convert the pressure from atm to Pa. Using the given number of molecules, calculate the number of moles of a gas. Use an ideal gas equation to determine the volume of the gas.

Formula used:
Ideal gas equation,
PV=nRTPV = nRT
Here, P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant and T is the temperature.

Complete step by step solution:
We have given the number of molecules, N=4×1024N = 4 \times {10^{24}}, temperature T=50KT = 50\,{\text{K}} and pressure of the gas P=0.03atmP = 0.03\,{\text{atm}}. Therefore, we can surely use the ideal gas equation to determine the volume of the vessel.
To use the ideal gas equation, we know that the pressure should be in Pascal. Therefore, we can convert the pressure from atm to Pa as follows,
1atm=105Pa1\,{\text{atm}} = {10^5}\,{\text{Pa}}
0.03atm=0.03×105Pa\Rightarrow 0.03{\text{atm}} = 0.03 \times {10^5}\,{\text{Pa}}
0.03atm=3000Pa\Rightarrow 0.03{\text{atm}} = 3000\,{\text{Pa}}
Also, we have given the number of molecules of the gas is N=4×1024N = 4 \times {10^{24}}. In ideal gas equation, n represents the number of moles of a gas. Therefore, we can calculate the number of moles of a gas as follows,
n=NNAn = \dfrac{N}{{{N_A}}}
Here, NA{N_A} is Avogadro’s number.
Substituting 6.022×10236.022 \times {10^{23}} for NA{N_A} and 4×10244 \times {10^{24}} for N in the above equation, we get,
n=4×10246.022×1023n = \dfrac{{4 \times {{10}^{24}}}}{{6.022 \times {{10}^{23}}}}
n=6.62moles\Rightarrow n = 6.62\,{\text{moles}}
We have the ideal gas equation,
PV=nRTPV = nRT
V=nRTP\Rightarrow V = \dfrac{{nRT}}{P}
Here, V is the volume, n is the number of moles, R is the gas constant and T is the temperature.
Substituting 6.62 moles for n, 8.314Jmol1K18.314\,J\,mo{l^{ - 1}}{K^{ - 1}} for R, 50 K for T and 3000 Pa for P in the above equation, we get,
V=(6.62)(8.314)(50)3000V = \dfrac{{\left( {6.62} \right)\left( {8.314} \right)\left( {50} \right)}}{{3000}}
V=0.91m3\Rightarrow V = 0.91\,{{\text{m}}^{\text{3}}}

Therefore, the volume of the gas is 0.91m30.91\,{{\text{m}}^{\text{3}}}.

Note:
Students often get confused between the number of moles and number of molecules and use the number of molecules instead of the number of moles in the ideal gas equation. Remember, n in the ideal gas equation represents the number of moles of the gas. Also, to use the ideal gas equation, the pressure of the gas should be in Pascal.