Question #205151

A sample of helium gas takes up 15 L. After being compressed down to a volume of 8 L, the sample has a pressure of 4.3 atm. What was the original pressure?

Expert's answer

P_{1} = unknown

V_{1} = 15 L

P_{2} = 4.3 atm

V_{2} = 8 L

T_{1} = T_{2} = const

**Solution:**

Since the temperature and amount of gas remain unchanged, Boyle's law can be used.

Boyle's gas law can be expressed as: P_{1}V_{1} = P_{2}V_{2}

To find the original pressure, solve the equation for P_{1}:

P_{1} = P_{2}V_{2 }/ V_{1}

P_{1} = (4.3 atm × 8 L) / (15 L) = 2.29 atm = 2.3 atm

P_{1} = 2.3 atm

**Answer: 2.3 atm is the original pressure.**

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