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Question: The pressure exerted by 6.0 gm of methane gas in a \(0.03 m^3\) vessel at \(129^oC\) is: (Atomic mas...

The pressure exerted by 6.0 gm of methane gas in a 0.03m30.03 m^3 vessel at 129oC129^oC is: (Atomic masses of C=12.01,H=1.01C = 12.01,H = 1.01 andR= 8.314JK−1mol−1R = {\text{ }}8.314J{K^{ - 1}}mo{l^{ - 1}})

A.215216 Pa B.13409 Pa C.41648 Pa D.31684 Pa  A.215216{\text{ }}Pa \\\ B.13409{\text{ }}Pa \\\ C.41648{\text{ }}Pa \\\ D.31684{\text{ }}Pa \\\
Explanation

Solution

Here, we are going to use the following ideal gas law equation to solve this problem.
PV=nRTPV = 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 gm6{\text{ }}gm
T = Temperature = 129oC+273=402K{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 = (12.01×4 )×1.01=16.05g  M{\text{ }} = {\text{ }}(12.01 \times 4{\text{ }}) \times 1.01 = 16.05g\;
Now, putting these values in ideal gas equation, we get
PV = nRTPV{\text{ }} = {\text{ }}nRT
PV=mass given (W) molar mass (M)RTPV = \dfrac{{{\text{mass given (W) }}}}{{{\text{molar mass (M)}}}}RT​ …………….. (As n = mass of gas/molar mass of gas)
∴P = mass given (W) molar mass (M)RTV{\text{P = }}\dfrac{{{\text{mass given (W) }}}}{{{\text{molar mass (M)}}}}\dfrac{{{\text{RT}}}}{{\text{V}}}
P = 6 ×8.314×402 16.05×0.03{\text{P = }}\dfrac{{{\text{6 }} \times 8.314 \times 402{\text{ }}}}{{16.05 \times 0.03}}
P = 20053.37 0.4815{\text{P = }}\dfrac{{20053.37{\text{ }}}}{{0.4815}}
P = 41647.70 PascalP{\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.