1.22g of a gas mesured over water at 15^(@)C and undr a pressure of 1.02 bar of mercury occupied 0.9 dm^(3). Calculate the volume of the volume of the dry gas at N.T.P. Vapour pressure of water at 15^(@)C is 0.018 bar.

1.22g of a gas mesured over water at 15^(@)C and undr a pressure of 1.

1.	1.22g of a gas mesured over water at 15^(@)C and undr a pressure of 1.02 bar of mercury occupied 0.9 dm^(3). Calculate the volume of the volume of the dry gas at N.T.P. Vapour pressure of water at 15^(@)C is 0.018 bar.
Answer» <html><body><p></p>Solution :Pressure of dry gas = Pressure of moist gas - <a href="https://interviewquestions.tuteehub.com/tag/aqueous-2440179" style="font-weight:bold;" target="_blank" title="Click to know more about AQUEOUS">AQUEOUS</a> tension <br/> `= <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.02 - 0.018 = 1.002` bar <br/> From the available data: `V_(1) = 0.9 dm^(3) , V_(2) = ?` <br/> `P_(1) = 1.002` bar , `P_(2) = 1.013` bar <br/> `T_(1) = 15+273 = <a href="https://interviewquestions.tuteehub.com/tag/288-1834218" style="font-weight:bold;" target="_blank" title="Click to know more about 288">288</a> K , T_(2) = 273 K` <br/> According the Gas equation , `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))` <br/> <a href="https://interviewquestions.tuteehub.com/tag/substituting-1231652" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTITUTING">SUBSTITUTING</a> the values, `V_(2) = ((1.022 "bar")xx (0.9 dm^(3)) xx (273 K))/((288 K) xx (1.013 "bar" )) = 0.844 dm^(3)`</body></html>