At a pressure of 405 kPa and a temperature of 200. K, the volume of a gas is 6.00 mL. At what pressure will the new temperature and volume be 300. K and 4.00 mL?
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Given,
Pressure 1 (P₁) = 405 kPa
= 405000 Pa
Volume 1 (V₁) = 6 mL
Temperature 1 (T₁) = 200 K
Volume 2 (V₂) = 4 mL
Temperature 2 (T₂) = 300 K
According to Ideal Gas Equation,
P₁V₁/T₁ = P₂V₂/T₂
= (405000*6)/200 = (P₂*4)/300
= 2430000/200 = 4P₂/300
= 12150 = 4P₂/300
= 3645000 = 4P₂
= 911250 Pa = P₂
= 911.25 kPa = P₂
The New Pressure will be 911.25 kPa.
I hope this helps.
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