Answer:
Explanation:
Apply Boyle's law of gas
P₁ V₁ = P₂ V₂ where P₁ and P₂ are initial and final pressure , V₁ , V₂ are initial and final volume .
1.00 atm x 3 liter = 3.60 atm x V₂
V₂ = 0.833 liter.
b )
Work done on the gas by piston = 2.303 RT log P₂ / P₁
= 2.303 x 8.3 x 300 K x log 3.6 atm / 1 atm
= 3190 J .
c )
Q = ΔE + W
Q = 0 + 3190 J
Heat energy transferred out Q = 3190 J
d )
Power = work done / time
= 3190 J / .020 second
= 159500 W .
159.5 kW.
Answer:
(a) 0.833 L
(b) - 383.32J
(c) - 383.32 J
(d) 19166.2 W
Explanation:
initial volume, V = 3 L
initial temperature, T = 300 K
Initial pressure, P = 1 atm
final pressure, P' = 3.6 atm
Temperature is constant.
(a) Let the final volume is V'.
As the temperature is constant,
P V = P' V'
1 x 3 = 3.6 x V'
V' = 0.833 L
(b) Let the number of moles is n
[tex]P V = n R T \\\\1\times 1.01 \times 10^5\times 3\times 10^{-3} = n\times 8.31\times 300\\\\n = 0.12[/tex]
The work done in isothermal process is
[tex]W = n T T lon{\frac{V'}{V}}\\\\W = 0.12 \times 8.31\times 300\times ln {\frac{0.833}{3}}\\\\W = - 383.32 J[/tex]
(c) The energy is given by the first law of thermodynamics
dQ = dU + dW
Here, dU is the zero as the temperature is constant.
So, the heat energy is
dQ = dW = - 383.32 J
(d) Time, t = 20 ms
Power is
[tex]P = \frac{W}{t} \\\\P = \frac{383.32}{0.02} =19166.2 W[/tex]
