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omeli [17]
3 years ago
12

A rigid tank whose volume is unknown is divided into two parts by a partition. One side of the tank contains an ideal gas at 935

°C. The other side is evacuated and has a volume twice the size of the part containing the gas. The partition is now removed and the gas expands to fill the entire tank. Heat is now transferred to the gas until the pressure equals the initial pressure. Determine the final temperature of the gas.
Physics
1 answer:
a_sh-v [17]3 years ago
7 0

Answer:

2805 °C

Explanation:

If the gas in the tank behaves as ideal gas at the start and end of the process. We can use the following equation:

P=RTn/V

The key issue is identify the quantities (P,T, V, n) in the initial and final state, particularly the quantities that change.

In the initial situation the gas have an initial volume V_{i}, temperature T_{i}, and pressure P,.

And in the final situation the gas have different volume V_{f} and temeperature T_{f}, the same pressure P,, and the same number of moles n,.

We can write the gas ideal equation for each state:

P=RT_{i}n/V_{i} and P=RT_{f}n/V_{f}, as the pressure are equals in both states we can write

RT_{i}n/V_{i} = RT_{f}n/V_{f}

solving for T_{f}

T_{f} = T_{i}/V_{i} * V_{f} (*)

We know T_{i}  = 935 °C, and that the V_{f} (the complete volume of the tank) is the initial volume V_{i} plus the part initially without gas which has a volume twice the size of the initial volume (read in the statement: the other side has a volume twice the size of the part containing the gas). So the final volume  V_{f}= V_{i} + 2V_{i}=3V_{i}

Replacing in (*)

T_{f} = 935/V_{i} * 3V_{i} = 935*3= 2805

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The length of an object is 10 cm. It means that 10 shows the magnitude of length and cm shows its unit.

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A sample of an unknown substance has a mass of 0. 158 kg. If 2,510. 0 J of heat is required to heat the substance from 32. 0°C
guapka [62]

The specific heat of the unknown substance with a mass of 0.158kg is 0.5478 J/g°C

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

The specific heat capacity of a substance can be calculated using the following formula:

Q = m × c × ∆T

Where;

  • Q = quantity of heat absorbed (J)
  • c = specific heat capacity (4.18 J/g°C)
  • m = mass of substance
  • ∆T = change in temperature (°C)

According to this question, 2,510.0 J of heat is required to heat the 0.158kg substance from 32.0°C to 61.0°C. The specific heat capacity can be calculated:

2510 = 158 × c × (61°C - 32°C)

2510 = 4582c

c = 2510 ÷ 4582

c = 0.5478 J/g°C

Therefore, the specific heat capacity of the unknown substance that has a mass of 0.158 kg is 0.5478 J/g°C.

Learn more about specific heat capacity at: brainly.com/question/2530523

4 0
2 years ago
200-grams of computer chips with a specific heat of 0.3 kJ/kg·K are initially at 25°C. These chips are cooled by placement in 0.
balu736 [363]

Answer:

a. -0.01324 kJ/K,  b.  = 0.03233 kJ/K , c.  = 0.01909, Yes the process is possible

Explanation:

Heat transfer will occur between the chip and the surrounding fluid. Then, finally they will attain a common equilibrium temperature and heat transfer will stop. Now, if we assume that, after heat transfer, chip will attain the temperature of fluid, that is, -34 C,, So , to check whether this is possible

Amount of energy lost by the chip = m . c . (T(i) - T(f))

= 0.2 x 0.3 (25 + 34) = 3.54 KJ

Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

Note that heat transfer will change the internal energy of the fluid. Do not consider enthalpy change, as this is not a problem involving fluid flow in and out of the system

M[(μf+ x . μfg) - μf] = m(xμfg)

<u>Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)</u>

3.54 = 0.1 . X x 203.29

<u>x = 0.1741, which is the dryness fraction of fluid at the final state.</u>

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

0.1 x μfg= <u>20.329 kJ, which is much greater than 3.45 kJ</u>, therefore, it is certain that the state of fluid will be at -34 C only and at the saturation pressure of 69.56 KPa. So the chip will come to attain the temperature of -34 C.  

a. Write the equation for the change of entropy in the chips

ΔSchips = mchips . c . ln(T2/T1), where mc is the mass of chips, c is the specific heat of chips, T2 is the temperature at state 2 and T1 is the temperature at state 1

Substitute mc = 0.2 kg, c = 0.3kJ/kg.K, T1 = 25 + 273, T2 = -34 + 273

delSchips = 0.2 x 0.3 x ln [(-34+273)/ (25+273)]

= -0.01324 kJ/K

There fore the change in entropy of the chips is -0.01324 kJ/K

b. Entropy change of fluid R- 134a

ΔS2 = m[Sfinal – S initial]

= m[Sf + x . Sfg - Sf]

= 0.2 x (0.1741 x 0.92859)

= 0.03233 kJ/K

c. Calculate the total change in the entropy of the entire system

delS = delSchips + delSR -134a

= -0.01324 + 0.03233

= 0.01909

<u>Since the total change in entropy of the entire system is positive that exactly explains that the actual processes are happening in the direction of increase of entropy therefore, the process is possible.</u>

<u />

6 0
3 years ago
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