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Alexandra [31]
3 years ago
10

A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. (a) Calculate the work done when the gas expands isothermally against

a constant external pressure of 200 Torr until its volume has increased by 3.3 dm3. (b) Calculate the work that would be done if the same expansion occurred reversibly.
Physics
1 answer:
valkas [14]3 years ago
7 0

Answer:

(A) Work done will be 87.992 KJ

(B) Work done will be 167.4 KJ            

Explanation:

We have given mass of methane m = 4.5 gram = 0.0045 kg

Volume occupies V_1=12.7dm^3=12.7liters

And volume is increased by 3.3dm^3 so V_2=12.7+3.3=16liters

Temperature T = 310 K

Pressure is given as 200 Torr = 26664.5 Pa

(a) At constant pressure work done is given by

W=P(V_2-V_1)=26664.5\times (16-12.7)=87992.85J=87.992kj

(b) At reversible process work done is given by W=nRTln\frac{V_2}{V_1}

We have given mass = 4.5 gram

Molar mass of methane = 16

So number of moles n=\frac{mass\ in\ gram}{mol;ar\ mass}=\frac{4.5}{16}=0.28125

So work done W=0.28125\times 8.314\times 310ln\frac{16}{12.7}=167.4J

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Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
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Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • mass of the larger disk = M_2\ =\ 1.6\ kg
  • Radius of the larger disk =R_2\ =\ 5.0\ cm\ =\ 0.05\ m
  • mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding,\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)

Net torque on the smaller disk,

\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)

From eqn (1) and (2), we get,

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let 'a_2' be the acceleration of the block in the second case,

From the above expression,

\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2

5 0
3 years ago
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