Question

A 739 mL gas sample at STP is compressed to

a volume of 244 mL, and the temperature is

increased to 24◦C. What is the new pressure

of the gas in Pa?

Answer 1

The new pressure is 3.295 × 10⁵ Pa

Step-by-step explanation:

Given: Initial Volume: V₁ = 739 mL, Final Volume: V₂ = 244 mL,

At STP; Initial Temperature: T₁ = 273 K, Initial Pressure: P₁ = 10⁵ Pa,

Final Temperature: T₂ = 24°C = 24 + 273 = 297 K, (∵ 0°C = 273.15 K)

Final Pressure: P₂ = ?

According to the Combined Gas Law:

`(P_(1)V_(1))/(T_(1))= (P_(2)V_(2))/(T_(2))`

`\Rightarrow P_(2) = (P_(1)V_(1)T_(2))/(T_(1)V_(2))`

`\Rightarrow P_(2) = (10^(5) \: Pa* 739 \: mL* 297 \: K)/(273 \: K* 244 \: mL)= 3.295 * 10^(5) \: Pa`

Therefore, the new or final pressure of the given gas: P₂ = 3.295 × 10⁵ Pa