Answer:
P V = n R T ideal gas equation
P2 V2 / P1 V1 = T2 / T1
V2 / V1 = T2 / T1 * P1 / P2 = T2 P1 / (T1 P2)
V2 / V1 = (1.17 T1) / T1 * (P1 / .22 P1) assuming absolute temp as 1.17 P1
V2 / V1 = 1.17 / .22 = 5.32
V = 4/3 pi R^3 = 4/3 pi (D/2)^3 = 4/3 pi D^3 / 8 = pi D^3 / 6
V2 / V1 = D2^3 / D1^3
D2 = (V2 / V1 * D1^3)^1/3
D2 = 5.32^1/3 * D = 1.75 D (D1 = D)
The "missing" 4 kg may have escaped the scene of the fire in the
form of hot gases, and as particles of soot and ash that were seen
leaving in the form of "smoke".
Answer:
a) W=12166.20876 J
b) U= -12166.20876 J
Explanation:
No. of moles, n = 8.41
Change of temperature, ΔT = T1 - T2
= 395 - 279
= 116 K
For monatomic gas, γ = 5/3
γ -1 = 2 /3
Solution:
(a)
Work done,
plugging values we get

Ans:
12166.20876 J
Work done, W = + 12166.20876 J
(b)
From first law of thermodynamics, dQ = U + W
but, dQ = 0
( adiabatic process)
Hence, U = - W
= - 12166.20876 J
Ans:
Change in internal energy, U = - 12166.20876 J
1 mile. Is this a joke lol