The pressure exerted by 6.0 gm of methane gas in a $0.03 m^3$ vessel at $129^oC$ is: (Atomic masses of \[C = 12.01,H = 1.01\] and\[R = {\text{ }}8.314J{K^{ - 1}}mo{l^{ - 1}}\])
\[
A.215216{\text{ }}Pa \\
B.13409{\text{ }}Pa \\
C.41648{\text{ }}Pa \\
D.31684{\text{ }}Pa \\
\]
Answer
656.1k+ views
Hint: Here, we are going to use the following ideal gas law equation to solve this problem.
\[PV = nRT\],
Where,
P is pressure, V is volume, n is number of moles of ideal gas, R is gas constant and T is temperature.
Complete step by step answer:
Note that, for the ideal gas equation, the product of pressure and volume is directly proportional to the temperature. It means that if the temperature of the gas remains constant, pressure or volume may increase on condition that the complementary variable decreases. Now, this also means that if the temperature of the gas varies, then it may lead to a change in the variable of pressure or volume.
Now, given from the question,
W = Mass of methane =\[6{\text{ }}gm\]
T = Temperature = \[{129^o}C + 273 = 402K\]
M = Molar mass of methane
∴ M = (no. of C atom × Atomic mass of C) + (no. of H atom × Atomic mass of H)
∴ \[M{\text{ }} = {\text{ }}(12.01 \times 4{\text{ }}) \times 1.01 = 16.05g\;\]
Now, putting these values in ideal gas equation, we get
\[PV{\text{ }} = {\text{ }}nRT\]
\[PV = \dfrac{{{\text{mass given (W) }}}}{{{\text{molar mass (M)}}}}RT\] …………….. (As n = mass of gas/molar mass of gas)
∴\[{\text{P = }}\dfrac{{{\text{mass given (W) }}}}{{{\text{molar mass (M)}}}}\dfrac{{{\text{RT}}}}{{\text{V}}}\]
\[{\text{P = }}\dfrac{{{\text{6 }} \times 8.314 \times 402{\text{ }}}}{{16.05 \times 0.03}}\]
\[{\text{P = }}\dfrac{{20053.37{\text{ }}}}{{0.4815}}\]
\[P{\text{ }} = {\text{ }}41647.70{\text{ }}Pascal\]
Hence, the correct option is option C.
Note:
We must know that the ideal gas equation is an important tool that gives us a very good estimate of gases at high temperatures and low pressures. The ideal gas equation makes it possible to study the relationship between the non-constant properties of ideal gases (n, P, V, T) providing three of these properties remain fixed.
\[PV = nRT\],
Where,
P is pressure, V is volume, n is number of moles of ideal gas, R is gas constant and T is temperature.
Complete step by step answer:
Note that, for the ideal gas equation, the product of pressure and volume is directly proportional to the temperature. It means that if the temperature of the gas remains constant, pressure or volume may increase on condition that the complementary variable decreases. Now, this also means that if the temperature of the gas varies, then it may lead to a change in the variable of pressure or volume.
Now, given from the question,
W = Mass of methane =\[6{\text{ }}gm\]
T = Temperature = \[{129^o}C + 273 = 402K\]
M = Molar mass of methane
∴ M = (no. of C atom × Atomic mass of C) + (no. of H atom × Atomic mass of H)
∴ \[M{\text{ }} = {\text{ }}(12.01 \times 4{\text{ }}) \times 1.01 = 16.05g\;\]
Now, putting these values in ideal gas equation, we get
\[PV{\text{ }} = {\text{ }}nRT\]
\[PV = \dfrac{{{\text{mass given (W) }}}}{{{\text{molar mass (M)}}}}RT\] …………….. (As n = mass of gas/molar mass of gas)
∴\[{\text{P = }}\dfrac{{{\text{mass given (W) }}}}{{{\text{molar mass (M)}}}}\dfrac{{{\text{RT}}}}{{\text{V}}}\]
\[{\text{P = }}\dfrac{{{\text{6 }} \times 8.314 \times 402{\text{ }}}}{{16.05 \times 0.03}}\]
\[{\text{P = }}\dfrac{{20053.37{\text{ }}}}{{0.4815}}\]
\[P{\text{ }} = {\text{ }}41647.70{\text{ }}Pascal\]
Hence, the correct option is option C.
Note:
We must know that the ideal gas equation is an important tool that gives us a very good estimate of gases at high temperatures and low pressures. The ideal gas equation makes it possible to study the relationship between the non-constant properties of ideal gases (n, P, V, T) providing three of these properties remain fixed.
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