Answer:
1) 6.0 atm.
2) 2.066 atm.
Explanation:
- From the general law of ideal gases:
<em>PV = nRT.</em>
where, P is the pressure of the gas.
V is the volume of the container.
n is the no. of moles of the gas.
R is the general gas constant.
T is the temperature of the gas (K).
<em>1) What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperature is held constant?</em>
- At constant T and at two different (P, and V):
<em>P₁V₁ = P₂V₂.</em>
P₁ = 2.0 atm, V₁ = 150.0 mL.
P₂ = ??? atm, V₂ = 50.0 mL.
<em>∴ P₂ = P₁V₁/V₂</em> = (2.0 atm)(150.0 mL)/(50.0 mL) = <em>6.0 atm.</em>
<em>2. A sample of a gas in a rigid container at 30.0°C and 2.00 atm has its temperature increased to 40.0°C. What will be the new pressure?</em>
<em></em>
- Since the container is rigid, so it has constant V.
- At constant V and at two different (P, and T):
<em>P₁/T₁ = P₂/T₂.</em>
P₁ = 2.0 atm, T₁ = 30.0°C + 273 = 303 K.
P₂ = ??? atm, T₂ = 40.0°C + 273 = 313 K.
<em>∴ P₂ = P₁T₂/T₁ </em>= (2.0 atm)(313.0 K)/(303.0 K) =<em> 2.066 atm.</em>
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