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Softa [21]
4 years ago
5

A cylinder with a piston contains 0.250 mol of oxygen at2.40x105 Pa and 355 K. The oxygen may be treated as an ideal gas.The gas

first expands isobarically to twice its original volume. Itis then compressed isothermally back to its original volume, andfinally it is cooled isochorically to its original pressure.
(a)Show the series of processes on a pV diagram.
(b) Compute thetemperature during the isothermal compression.
(c) Compute themaximum pressure.
(d) Compute the total work done by the piston onthe gas during the series of processes.

Physics
1 answer:
Nikitich [7]4 years ago
7 0

Answer:

= 285 Joules

Explanation:

a) answer can be found out in attachment

(b) The temperature for the isothermal compression is the same as the temp at the end of the isobaric expansion. Since pressure is held constant but volume doubles, we use the ideal gas law:

p V = nR T          to see that the temperature also doubles.

.So...   temp for isothermal compression =   355×2 = 710 K

.(c)   The max pressure occurs at the top point. At this point, the volume is back to the original value but the temperature is twice the original value. So the pressure at this point is twice the original, or

max pressure = 2×240000 Pa =  480000 Pa  =   4.80 x 10^5 Pa

(d) total work done by the piston = workdone during isothermal compression - work done during expansion =

=  nRT ln(V initial / V final)-p (V initial - V final)

=   nRT ln(2) - nR(T final - T initial)  

= 0.250× 8.314 ×710×ln(2)-0.250×8.314× (710 - 355)

=     285 Joules

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True or false : In crystalline solids the particles are not arranged in a regular pattern
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Describe what happens when heat is removed from a system
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This happens<span> at the boiling </span>temperature<span> of every substance that can vaporize. At the boiling </span>temperature<span>, adding </span>heat<span> energy converts the liquid into a gas WITHOUT RAISING THE </span>TEMPERATURE<span>. Adding </span>heat<span> to a boiling liquid is an important exception to general rule that more </span>heat<span> makes a higher </span>temperature<span>.</span>
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Find the quantity of heat needed
krok68 [10]

Answer:

Approximately 3.99\times 10^{4}\; \rm J (assuming that the melting point of ice is 0\; \rm ^\circ C.)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}

The energy required comes in three parts:

  • Energy required to raise the temperature of that 0.100\; \rm kg of ice from (-10\; \rm ^\circ C) to 0\; \rm ^\circ C (the melting point of ice.)
  • Energy required to turn 0.100\; \rm kg of ice into water while temperature stayed constant.
  • Energy required to raise the temperature of that newly-formed 0.100\; \rm kg of water from 0\; \rm ^\circ C to 10\;\ rm ^\circ C.

The following equation gives the amount of energy Q required to raise the temperature of a sample of mass m and specific heat capacity c by \Delta T:

Q = c \cdot m \cdot \Delta T,

where

  • c is the specific heat capacity of the material,
  • m is the mass of the sample, and
  • \Delta T is the change in the temperature of this sample.

For the first part of energy input, c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}.

Similarly, for the third part of energy input, c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}.

The second part of energy input requires a different equation. The energy Q required to melt a sample of mass m and latent heat of fusion L_\text{f} is:

Q = m \cdot L_\text{f}.

Apply this equation to find the size of the second part of energy input:

\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}.

Find the sum of these three parts of energy:

\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}.

3 0
3 years ago
The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc
qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

R = \frac{6}{2.5} - 2.4 ohm

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

\frac{3R}{R+3} \times 2.5 = 6

R= 12 ohm

2) circuit Q

I\times (R+0.1) =V

R+0.1 =\frac{6}{2.5}

R = 2.3 ohm

5 0
3 years ago
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