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A container of volume 200 cm$^3$ contains 0.2 mole of hydrogen gas and 0.3 mole of argon gas. The pressure of the system at temperature 200 K ($R=8.3\ JK^{-1}mol^{-1}$) will be:-
A
$6.15\times10^5$ Pa
B
$6.15\times10^4$ Pa
C
$4.15\times10^5$ Pa
D
$4.15\times10^6$ Pa
Detailed Solution
$P_{mix}=\dfrac{(\mu_1+\mu_2)RT_{mix}}{V_{mix}}=\dfrac{(0.2+0.3)\times8.3\times200}{200\times10^{-6}}=\dfrac{0.5\times8.3\times200}{200\times10^{-6}}=4.15\times10^6$ Pa.