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Exercises · 5.5

Q.Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

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Step 1 – Set up for gas A alone

1 g of gas A in flask of volume VV at T=27°C=300 KT=27°\text{C}=300\ \text{K} gives pA=2p_A = 2 bar:

pAV=nART=1MART...(i)p_A V = n_A RT = \frac{1}{M_A}RT \quad \text{...(i)}

Step 2 – Set up for gas B added to the same flask

After adding 2 g of gas B (same VV, same TT), total pressure becomes 3 bar. By Dalton's law, the partial pressure contributed by B is:

pB=ptotal−pA=3−2=1 barp_B = p_{\text{total}} - p_A = 3 - 2 = 1\ \text{bar}

pBV=nBRT=2MBRT...(ii)p_B V = n_B RT = \frac{2}{M_B}RT \quad \text{...(ii)}

Step 3 – Divide (i) by (ii) — VV, RR, TT all cancel …

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