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elena-s [515]
3 years ago
9

In a standard gas grill propane tank, there is approximately 4,579 mL of propane (C3H8). At a temperature of 55˚C, the tank has

a pressure of 1,798 kPa. The tank is cooled to 25˚C and the pressure reduced to 1,025 kPa. What will the new volume be?
Physics
1 answer:
AnnyKZ [126]3 years ago
8 0

Answer:

7300 mL

Explanation:

The question is a bit misleading, since tanks generally have a constant volume.  But anyway, using ideal gas law:

PV = nRT

If amount of gas (n) is constant, then:

PV / T = constant

P₁V₁ / T₁ = P₂V₂ / T₂

(1798 kPa) (4579 mL) / (55 + 273) K = (1025 kPa) V / (25 + 273) K

V = 7298 mL

Rounded to two significant figures, the volume is 7300 mL.

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(a)0.531m/s

(b)0.00169

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Mass of block 2,m_2=1626 g=1.626 kg

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Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

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k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

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