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3 years ago

Change in enthalpy when 11.2 dm3 of helium at ntp is heated in a cylinder to 100 degrees celsius

Chemistry

1 answer:

REY [17]3 years ago

8 0

Answer:

Enthalpy change = 333.3 J

Explanation:

Given data:

Volume of helium = 11.2 dm³ or 11.2 L

Temperature = 100 °C (100+273= 373 K)

Enthalpy change = ?

Solution:

Normal temperature of room = 20°C (20+273 = 293 K)

one mole of gas occupy volume = 22.4 L

ΔT = final temperature - initial temperature

ΔT = 373K - 293K

ΔT = 80K

Number of moles of gas = 11.2 L/22.4 l

Number of moles of gas = 0.5 mol

Enthalpy change = nRΔT

Enthalpy change = 0.5 mol × 8.3j/mol.k ×80 K

Enthalpy change = 333.3 J

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- How much power does it take to lift a<br> 1,000 N load 10 m in 20 s?

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Answer:

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Explanation:

The given values are:

Force,

F = 1000 N

Velocity,

s = 10 m

Time,

t = 20 s

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The power will be:

= $\frac{Force\times Velocity}{Time}$

On putting the values, we get

= $\frac{1000\times 10}{20}$

= $\frac{10000}{20}$

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