Estimate the max vapor concentration that could be expected in the head space of a 55 gallon drum containing n-Pryopyl nitrate with VP of 20?

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Multiple Choice

Estimate the max vapor concentration that could be expected in the head space of a 55 gallon drum containing n-Pryopyl nitrate with VP of 20?

Explanation:
When a liquid sits in a closed container, the vapor in the headspace reaches equilibrium with the liquid and the vapor’s partial pressure equals the liquid’s vapor pressure. In air at about atmospheric pressure (roughly 760 mmHg), the mole fraction of the vapor is P_vap / P_total. With a vapor pressure of 20 mmHg, the fraction is 20/760 ≈ 0.0263, which is about 2.63% by volume. Converting to parts per million gives roughly 26,000 ppm. The size of the drum doesn’t change this maximum concentration, because at equilibrium the headspace composition is determined by the vapor pressure and total pressure, provided there’s enough liquid to maintain the vapor pressure. Temperature matters, since vapor pressure changes with temperature, so the 20 mmHg value must correspond to the temperature of interest. So the estimated max concentration is about 26,000 ppm.

When a liquid sits in a closed container, the vapor in the headspace reaches equilibrium with the liquid and the vapor’s partial pressure equals the liquid’s vapor pressure. In air at about atmospheric pressure (roughly 760 mmHg), the mole fraction of the vapor is P_vap / P_total. With a vapor pressure of 20 mmHg, the fraction is 20/760 ≈ 0.0263, which is about 2.63% by volume. Converting to parts per million gives roughly 26,000 ppm. The size of the drum doesn’t change this maximum concentration, because at equilibrium the headspace composition is determined by the vapor pressure and total pressure, provided there’s enough liquid to maintain the vapor pressure. Temperature matters, since vapor pressure changes with temperature, so the 20 mmHg value must correspond to the temperature of interest. So the estimated max concentration is about 26,000 ppm.

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