QUESTION IMAGE
Question
chemistry 20
final exam 2025
name: ebteasam
gases
use the following information to answer the next question.
a sample of helium gas occupies 465 ml at 400 kpa. more pressure is applied from
the surroundings and the final pressure becomes 4.94 atm, but the temperature
remains unchanged.
- the final volume occupied by the gas will be
□ 320 ml
□ 421 ml
□ 372 ml
□ 475 ml
Step1: Convert pressure units
We know that \(1\space atm = 101.325\space kPa\). So, \(P_2=4.9\space atm\times101.325\space kPa/atm = 496.4925\space kPa\), \(P_1 = 400\space kPa\), \(V_1=465\space mL\).
Step2: Apply Boyle's Law
Boyle's Law is \(P_1V_1=P_2V_2\) (since \(T\) is constant). Then \(V_2=\frac{P_1V_1}{P_2}\).
Substitute the values: \(V_2=\frac{400\space kPa\times465\space mL}{496.4925\space kPa}\)
\(V_2=\frac{186000}{496.4925}\space mL\approx 375\space mL\approx372\space mL\) (approximate due to rounding in pressure - unit conversion)
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\(372\space mL\)