Question

When 6.0 L of He(g) and 10. L of N2(g), both at 0oC and 1.0 atm, are pumped into an evacuated 4.0 L rigid container, the final pressure in the container at 0oC is

Answer 1

Answer:

P2=4atm

Explanation:

When 6.0 L of He(g) and 10. L of N2(g), both at 0oC and 1.0 atm, are pumped into an evacuated 4.0 L rigid container, the final pressure in the container at 0oC is

from boyle's law which states that the pressure on a gieben mass of gas is inversely proportional to the volume provided that temperature is kept constant

P∝1/v

PV=k

where k is the constant of proportionality

$P_{1} V_{1}=P_{2}V_{2}$

the v1=6L+10L=16L

P1=1atm

P2=?

V2=4L

16*1=4*P2

P2=4atm

P1=pressure 1atm

V1=volume of helium with nitrogen gas

P2=final pressure

V2=4l volume space

Answer 2

Answer:

The final pressure in the container is 4 atm.

Explanation:

Given data:

The volume of helium is, $V_{h}=6.0 \;\rm L$.

The volume of nitrogen is, $V_{n}=10.0 \;\rm L$.

Volume of container is, $V_{c}=4.0 \;\rm L$.

Applying the ideal gas equation for helium at standard temperature and pressure as,

$PV_{h}=n_{h}RT................................................................(1)$        

Here, P is the standard pressure (1 atm), $n_{h}$ is the number of moles of helium, R is the universal gas constant (0.082 L-atm/mol-K) and T is the temperature (273 K).

Solving equation (1) as,

$1 \times 6.0=n_{h} \times 0.082 \times 273\\n_{h} = 0.27 \;\rm moles$

Applying the ideal gas equation for nitrogen at standard temperature and pressure as,

$PV_{n}=n_{n}RT\\1 \times 10 =n_{n} \times 0.082 \times 273\\n_{n} = 0.45 moles$

Total number of moles is,

$N=n_{h}+n_{n}\\N=0.27+0.45 = 0.72 moles$

Final pressure by ideal gas equation is,

$P_{final} \times V_{c}=NRT\\P_{final} \times 4=0.72 \times 0.082 \times 273\\P_{final}=\frac{0.72 \times 0.082 \times 273}{4}\\ P_{final} \approx 4 \;\rm atm$

Thus, the final pressure in the container is 4 atm.

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