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NEET PHYSICSMedium

A container of volume 200 cm3200 \text{ cm}^3 contains 0.2 mole of hydrogen gas and 0.3 mole of argon gas. The pressure of the system at temperature 200 K200 \text{ K} (R=8.3 J K−1 mol−1R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}) will be:

A

6.15 × 10⁵ Pa

B

6.15 × 10⁴ Pa

C

4.15 × 10⁵ Pa

D

4.15 × 10⁶ Pa

Step-by-Step Solution

According to Dalton's Law of Partial Pressures, the total pressure exerted by a mixture of non-reactive gases is the sum of the partial pressures of individual gases. Alternatively, using the Ideal Gas Equation PV=nRTPV = nRT for the total mixture:

  1. Calculate Total Moles (nn): n=nH2+nAr=0.2+0.3=0.5 moln = n_{H_2} + n_{Ar} = 0.2 + 0.3 = 0.5 \text{ mol}

  2. Convert Volume to SI Units (VV): V=200 cm3=200×10−6 m3=2×10−4 m3V = 200 \text{ cm}^3 = 200 \times 10^{-6} \text{ m}^3 = 2 \times 10^{-4} \text{ m}^3

  3. Apply Ideal Gas Equation: P=nRTVP = \frac{nRT}{V} P=0.5×8.3×2002×10−4P = \frac{0.5 \times 8.3 \times 200}{2 \times 10^{-4}} P=8302×10−4P = \frac{830}{2 \times 10^{-4}} P=415×104 PaP = 415 \times 10^4 \text{ Pa} P=4.15×106 PaP = 4.15 \times 10^6 \text{ Pa}

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