Answer:
The new pressure is 3.295 × 10⁵ Pa
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:
[tex]\frac{P_{1}V_{1}}{T_{1}}= \frac{P_{2}V_{2}}{T_{2}}[/tex]
[tex]\Rightarrow P_{2} = \frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}[/tex]
[tex]\Rightarrow P_{2} = \frac{10^{5} \: Pa\times 739 \: mL\times 297 \: K}{273 \: K\times 244 \: mL}= 3.295 \times 10^{5} \: Pa[/tex]
Therefore, the new or final pressure of the given gas: P₂ = 3.295 × 10⁵ Pa
