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Alex_Xolod [135]

3 years ago

5

The volume of an ideal gas is adiabatically reduced from 151 L to 80.6 L. The initial pressure and temperature are 1.50 atm and

260 K. The final pressure is 3.61 atm. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is the final temperature? (c) How many moles are in the gas?

1 answer:

Zolol [24]3 years ago

4 0

Answer:

gas is dioatomic

T_f = 330.0 K

$\eta = 7.07 mole$

Explanation:

Part 1

below equation is used to determine the type Gas by determining $\gamma$ value

$\frac{V_{1}}{V_{F}}\gamma=\frac{P_{i}}{P_{f}}$

where V_i and V_f is initial and final volume respectively

and P_i and P_f are initial and final pressure

$\gamma = \frac{ln(P_f/P_i)}{ln(V_i/V_f)}$

$\gamma = \frac{ln(3.61/1.50)}{ln(151/80.6}$

\gamma = 1.38

therefore gas is dioatomic

Part 2

final temperature in adiabatic process is given as

$T_f = T_i*[\frac{v_i}{V_f}](^\gamma-1)$

substituing value to get final temperature

$T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}$

T_f = 330.0 K

Part 3

determine number of moles by using following formula

$\eta =\frac{PV}{RT}$

$\eta =\frac{1.013*10^{5}*0.151}{8.314*260}$

$\eta = 7.07 mole$

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Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m $V_p$ + 14m $V_n$

then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

$K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2$

and

Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

270 = $V_p$ + $14V_n$ (eq. 1)

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

36450 = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$ (eq. 2)

from this last equation we solve for $V_n$ as:

$V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

and replace in the other equation as:

270 = $V_p +$ 14 $\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

so,

$V_p = -267.258 m/s$

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

$V_n = 38.375 m/s$

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