A sealed container with a movable piston holds 0.80 mol of an ideal gas at an initial pressure of 1.2 × 10^5 Pa and an initial volume of 0.020 m^3. The gas is compressed isothermally until its volume is 0.012 m^3. What is the final pressure of the gas?
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Correct answer
B. 2.0 × 10^5 Pa
Principle or equation
For an isothermal process of an ideal gas, the ideal gas law PV = nRT reduces to P1V1 = P2V2 when n and T are constant.
Why this answer is correct
Since the temperature and amount of gas remain constant, apply Boyle's law: P1V1 = P2V2. Substitute P1 = 1.2 × 10^5 Pa, V1 = 0.020 m^3, V2 = 0.012 m^3. Solve for P2 = P1V1/V2 = (1.2 × 10^5)(0.020)/(0.012) = 2.0 × 10^5 Pa.
Example
If 1.0 m^3 of gas at 1.0 × 10^5 Pa is compressed to 0.50 m^3 isothermally, the new pressure is (1.0 × 10^5)(1.0)/(0.50) = 2.0 × 10^5 Pa.
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