A 200.0-mL sample of helium gas is in a cylinder with a movable piston at 110 kPa and 275 K. The piston is pushed in until the sample has a volume of 95.0 mL. The new temperature of the gas is 310. K. What is the new pressure of the sample? 51.1 kPa
261 kPa
249 kPa
274 kPa

P₂ = 261 KPa

Explanation:

Given data:

Volume of sample = 200 mL

Initial pressure = 110 KPa

Initial temperature = 275 K

Final volume = 95 mL

Final temperature = 310 K

Final pressure = ?

Solution:

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁V₁T₂ /T₁ V₂

P₂ = 110 KPa× 200 mL× 310 K /275 K  × 95 mL

P₂ = 6820000 KPa / 26125

P₂ = 261 KPa

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atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer