1. Consider two containers with the same volume and temperature. Container One holds “dry” air—a mixture of nitrogen and oxygen. Container Two holds “moist” air. The “moist” air has the same ratio of nitrogen to oxygen molecules, but also contains water vapor. According to the ideal gas law, if the pressures are equal, the weight of the gas in Container One will be:

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Here is the answer for the question – 1. Consider two containers with the same volume and temperature. Container One holds “dry” air—a mixture of nitrogen and oxygen. Container Two holds “moist” air. The “moist” air has the same ratio of nitrogen to oxygen molecules, but also contains water vapor. According to the ideal gas law, if the pressures are equal, the weight of the gas in Container One will be:. You’ll find the correct answer below

1. Consider two containers with the same volume and temperature. Container One holds “dry” air—a mixture of nitrogen and oxygen. Container Two holds “moist” air. The “moist” air has the same ratio of nitrogen to oxygen molecules, but also contains water vapor. According to the ideal gas law, if the pressures are equal, the weight of the gas in Container One will be:

The Correct Answer is

heavier than the gas inside the second container

Reason Explained

heavier than the gas inside the second container is correct for 1. Consider two containers with the same volume and temperature. Container One holds “dry” air—a mixture of nitrogen and oxygen. Container Two holds “moist” air. The “moist” air has the same ratio of nitrogen to oxygen molecules, but also contains water vapor. According to the ideal gas law, if the pressures are equal, the weight of the gas in Container One will be:

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