Answer:
पत्र- अपने क्षेत्र की सड़कों की बुरी दशा की जानकारी देते हुए और उन्हें ठीक कराने की प्रार्थना करते हुए नगर निगम अधिकारी को पत्र लिखिए |
Answer:
1. K.E = 11.2239 kJ ≈ 11.224 kJ
2. ![C_{V} = 37.413 JK^{- 1}](https://tex.z-dn.net/?f=C_%7BV%7D%20%3D%2037.413%20JK%5E%7B-%201%7D)
3. ![Q = 10.7749 kJ](https://tex.z-dn.net/?f=Q%20%3D%2010.7749%20kJ)
Solution:
Now, the kinetic energy of an ideal gas per mole is given by:
K.E = ![\frac{3}{2}mRT](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B2%7DmRT)
where
m = no. of moles = 3
R = Rydberg's constant = 8.314 J/mol.K
Temperature, T = 300 K
Therefore,
K.E = ![\frac{3}{2}\times 3\times 8.314\times 300](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B2%7D%5Ctimes%203%5Ctimes%208.314%5Ctimes%20300)
K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ
Now,
The heat capacity at constant volume is:
![C_{V} = \frac{3}{2}mR](https://tex.z-dn.net/?f=C_%7BV%7D%20%3D%20%5Cfrac%7B3%7D%7B2%7DmR)
![C_{V} = \frac{3}{2}\times 3\times 8.314 = 37.413 JK^{- 1}](https://tex.z-dn.net/?f=C_%7BV%7D%20%3D%20%5Cfrac%7B3%7D%7B2%7D%5Ctimes%203%5Ctimes%208.314%20%3D%2037.413%20JK%5E%7B-%201%7D)
Now,
Required heat transfer to raise the temperature by
is:
![Q = C_{V}\Delta T](https://tex.z-dn.net/?f=Q%20%3D%20C_%7BV%7D%5CDelta%20T)
![\Delta T = 15^{\circ} = 273 + 15 = 288 K](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%2015%5E%7B%5Ccirc%7D%20%3D%20273%20%2B%2015%20%3D%20288%20K)
![Q = 37.413\times 288 = 10774.9 J = 10.7749 kJ](https://tex.z-dn.net/?f=Q%20%3D%2037.413%5Ctimes%20288%20%3D%2010774.9%20J%20%3D%2010.7749%20kJ)
Not at all. Density on its own is not sufficient
Answer:
ΔV = -2.1 L
Explanation:
To solve this exercise we can use the ideal gas equation for two points
PV = nRT
P₁V₁ = P₂ V₂
where point 1 is on the surface and point 2 is at the desired depth,
V₂ =
let's calculate
V₂ = (
) 3.5 L
V₂ = 1.4 L
this is the new volume, the change in volume is
ΔV = V₂ -V₁
ΔV = 1.4-3.5
ΔV = -2.1 L
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I think hope you find this helpful.