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Difficulty: HardGas Laws and the Ideal Gas Equation

A gas cylinder fitted with a frictionless piston contains a fixed mass of ideal gas occupying a volume of 0.040 m30.040\text{ m}^3 at a pressure of 1.50×105 Pa1.50 \times 10^5\text{ Pa} and a temperature of 27∘C27^\circ\text{C}. The gas is heated to 127∘C127^\circ\text{C} while expanding to a new volume of 0.080 m30.080\text{ m}^3. What is the final pressure of the gas?

  1. 1.00×105 Pa1.00 \times 10^5\text{ Pa}Answer
  2. B
    3.53×105 Pa3.53 \times 10^5\text{ Pa}
  3. C
    7.50×104 Pa7.50 \times 10^4\text{ Pa}
  4. D
    4.00×105 Pa4.00 \times 10^5\text{ Pa}

Answer

The final pressure of the gas is 1.00×105 Pa1.00 \times 10^5\text{ Pa}.
By converting temperatures to absolute zero scale (300 K300\text{ K} and 400 K400\text{ K}) and applying the Combined Gas Law P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}, the final pressure P2P_2 evaluates directly to 1.00×105 Pa1.00 \times 10^5\text{ Pa}.

Step-by-Step Solution

1
Convert temperatures from Celsius to Kelvin.
T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K} and T2=127+273=400 KT_2 = 127 + 273 = 400\text{ K}.
Gas laws require absolute temperatures measured on the Kelvin scale.
2
Apply the Combined Gas Law formula P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}.
(1.50×105 Pa)(0.040 m3)300 K=P2(0.080 m3)400 K\frac{(1.50 \times 10^5\text{ Pa})(0.040\text{ m}^3)}{300\text{ K}} = \frac{P_2 (0.080\text{ m}^3)}{400\text{ K}}.
The mass of the gas is fixed while pressure, volume, and temperature all change.
3
Rearrange and solve for final pressure P2P_2.
P2=1.50×105×(0.0400.080)×(400300)=1.00×105 PaP_2 = 1.50 \times 10^5 \times \left(\frac{0.040}{0.080}\right) \times \left(\frac{400}{300}\right) = 1.00 \times 10^5\text{ Pa}.
Simplifying the numerical expression yields the final equilibrium pressure.

Key Concept

Combined Gas Law
Estimated Time:2m 0s
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Topics

Gas Laws and the Ideal Gas Equation

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