Question 998: Thermodynamics & KTG

Question

Two gases occupy two containers $$A$$ and $$B$$. The gas in $$A$$ of volume $$0.20 \mathrm{~m}^3$$, exerts a pressure of $$1.40 \mathrm{~MPa}$$ and that in $$B$$, of volume $$0.30 \mathrm{~m}^3$$ exerts a pressure of $$0.7 \mathrm{~MPa}$$. The two containers and united by a tube of negligible volume and the gases are allowed to exchange. Then, if the temperature remains constants. the final pressure in the container will be (in MPa).

Accepted Answer

To find the final pressure in the container when the two gases are allowed to mix, we can use the law of partial pressures. The total pressure in the combined volume is the sum of the partial pressures of the gases. However, since the temperatures are constant and we're dealing with ideal gases, we can also use the ideal gas law given by the equation $$PV = nRT$$, where $$P$$ stands for pressure, $$V$$ for volume, $$n$$ for the number of moles, $$R$$ for the ideal gas constant, and $$T$$ for temperature. Since the temperature and the gas constant do not change, and the system is closed (no gas is added or removed), we can simplify this situation using the formulas based on the initial and final conditions.
Given:
Container $$A$$: Volume ($$V_A$$) = $$0.20 \, 	ext{m}^3$$, Pressure ($$P_A$$) = $$1.40 \, 	ext{MPa}$$
Container $$B$$: Volume ($$V_B$$) = $$0.30 \, 	ext{m}^3$$, Pressure ($$P_B$$) = $$0.70 \, 	ext{MPa}$$
First, let us convert pressures to a consistent unit, but here since both are in MPa, and we expect our answer in MPa, we can keep them as is.
The total volume when the gases are allowed to mix is the sum of their individual volumes:
$$V_{	ext{total}} = V_A + V_B = 0.20 \, 	ext{m}^3 + 0.30 \, 	ext{m}^3 = 0.50 \, 	ext{m}^3$$
To find the final pressure, $$P_{	ext{final}}$$, we can use the conservation of the amount of gas since the temperature and the amount of gas are constant. This is reflected in the equation:
$$P_1 V_1 + P_2 V_2 = P_{	ext{final}}(V_1 + V_2)$$
Substituting the given values:
$$1.40 	imes 0.20 + 0.70 	imes 0.30 = P_{	ext{final}} 	imes 0.50$$
$$0.28 + 0.21 = P_{	ext{final}} 	imes 0.50$$
$$0.49 = P_{	ext{final}} 	imes 0.50$$
To find $$P_{	ext{final}}$$, divide both sides by $$0.50 \, 	ext{m}^3$$:
$$P_{	ext{final}} = \frac{0.49}{0.50} = 0.98 \, 	ext{MPa}$$
Therefore, the final pressure in the container after the gases are mixed, keeping the temperature constant, is $$0.98 \, 	ext{MPa}$$. Thus, the correct answer is:
Option B: 0.98

Answer

0.70

Answer

0.98

Answer

1.40

Answer

2.1

Thermodynamics & KTG

Choose the correct answer: