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pickupchik [31]
3 years ago
14

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 p

ressure in the container at 0oC is
Physics
1 answer:
Murrr4er [49]3 years ago
6 0

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

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A thermos contains m1 = 0.89 kg of tea at T1 = 31° C. Ice (m2 = 0.075 kg, T2 = 0° C) is added to it. The heat capacity of both w
lorasvet [3.4K]

Answer:

a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

b) T = 295.37 K

Explanation:

Given;

Initial temperature of tea T1 = 31 C

Initial temperature of ice T2 = 0 C

Mass of tea m1 = 0.89 kg

Mass of ice m2 = 0.075kg

The heat capacity of both water and tea c = 4186 J/(kg⋅K)

the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg

And T = the final temperature of the mixture

Heat loss by tea = heat gained by ice

m1c∆T1 = m2c∆T2 + m2Lf

m1c(T1-T) = m2c(T-T2) + m2Lf

m1cT1 - m1cT = m2cT - m2cT2 + m2Lf

m1cT + m2cT = m1cT1 + m2cT2 - m2Lf

T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

Substituting the values;

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)

T = 22.37 °C

T = 273 + 22.37 K

T = 295.37 K

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3 years ago
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