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Gaseous State question

2025 · Q57
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Gaseous State question

2025 · Q57

IIT JAMChemistryGaseous StateNumerical+2 / −0
In one second, 95 moles of He gas particles are hitting a wall of a cubic container of volume 1 dm 3^{3}3. If the average velocity component of the particles perpendicular to the wall is 1000 m s −1^{-1}−1, then the pressure of the gas in the container is X×105X \times 10^5X×105 N m −2^{-2}−2. The value of X is ‾\underline{\hspace{2cm}}​. (round off to two decimal places) [Given: Avogadro's number, NA=6.02×1023N_A = 6.02 \times 10^{23}NA​=6.02×1023, mass of He = 4 g mol −1^{-1}−1]
Numerical answer
View written solutionFree

Correct answer: 0.74, 0.78

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