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The pressure exerted by of methane gas in a vessel at (Atomic masses:$\text{C}=12\cdot 01,\text{ H}=1\cdot 01$ and$\text{R}=8\cdot 314\text{ J}{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{-1}}$).
A. 215216 Pa
B. 13409 Pa
C. 41684 Pa
D. 31684Pa

Last updated date: 18th Jun 2024
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Hint: In this question, temperature, volume and pressure given. The equation which gives the simultaneous effect of pressure and temperature on the volume of a gas is known as the ideal gas equation.

Formula used:
$ PV = nRT$

Complete step by step answer:
In the given statement, the weight of methane gas molecule is$6\cdot 0\text{g}$
                                                        $\text{W}=6\cdot 0\text{g}$ (given)
Volume of methane molecule$0\cdot 003\text{ }{{\text{m}}^{3}}$
                                                     $\text{V}=0\cdot 003\text{ }{{\text{m}}^{3}}$
Temperature $=129{}^\circ \text{C}$
Applying the ideal gas equation
PV = nRT
 P = Pressure
V = Volume
n = No. of moles
R = Gas constant$8\cdot 314\text{J}{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{-1}}$
T = Temperature
PV = nRT where,$\text{n}=\dfrac{\text{W}}{\text{M}}$
Here, W= weight of molecule and M = molecular mass
Molecular mass of methane molecule$=12\cdot 01+4\cdot 04=16\cdot 05$
Put all the values in equation
$\text{P}\times \text{0}\cdot 03=\dfrac{6\times 8\cdot 314\times 402}{16\cdot 05}$
$\text{P}=\dfrac{6\times 8\cdot 314\times 402}{16\cdot 05\times 0\cdot 003}$
On solving we get
$41647\cdot 7\text{ Pa}\cong \text{41648}$.

So, the correct answer is “Option C”.

 The terms in the ideal gas equation should be clear. The value of gas constant must be known .No gas in the actual condition shows the ideal behavior that is known as real gas.
In actual practice no gas behaves as ideal gas . All the gas shows deviation. There is change in pressure and volume of gas. An ideal gas equation changed into a real gas or Vander wall gas equation.
Where V is the volume of the gas, R is the universal gas constant, T is temperature, P is pressure, and V is volume. a and b are constant terms. When the volume V is large, b becomes negligible in comparison with V, a/V2 becomes negligible with respect to P, and the van der Waals equation reduces to the ideal gas law, PV=nRT.
$\left[ \text{P}+\dfrac{\text{a}{{\text{n}}^{2}}}{{{\text{v}}^{2}}} \right]\left( \text{V}-\text{nb} \right)=\text{nRT}$