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what pressure, in atm, is exerted by 2.50 l of gas containing 1.35 mol …

Question

what pressure, in atm, is exerted by 2.50 l of gas containing 1.35 mol at 320 k? pv = nrt

Explanation:

Step1: Rearrange the ideal gas law

The ideal gas law is \(PV = nRT\). Rearranging for pressure \(P\), we get \(P=\frac{nRT}{V}\).

Step2: Substitute the values

Given \(n = 1.35\space mol\), \(R=0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 320\space K\), and \(V=2.50\space L\).
Substitute into the formula: \(P=\frac{1.35\times0.0821\times320}{2.50}\).
First calculate the numerator: \(1.35\times0.0821\times320 = 1.35\times26.272=35.4672\).
Then divide by \(2.50\): \(P=\frac{35.4672}{2.50}=14.18688\approx14.19\space atm\).

Answer:

14.19 atm