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rodikova [14]
3 years ago
10

The pressure in an engine cylinder increases from 7.6 atm to 13 atm as the gas is compressed from 0.50 L to 0.10 L. What is the

expected cylinder temperature if the gas was initially 15°C?
Chemistry
2 answers:
REY [17]3 years ago
8 0
T k = 15 + 273 = 288 K

4.6 / 13 => 0.353 atm

0.50 / 0.10 => 5 L

<span>(15 + 273) K x (13 atm / 7.6 atm) x (0.50 L / 0.10 L) 
</span>
<span>= </span>2463.15 K

<span>hope this helps!</span>
Mkey [24]3 years ago
8 0

Answer:

98.53K or 174.47°C

Explanation:

Initial Pressure, P1 = 7.6atm

Final Pressure, P2 = 13atm

Initial Volume, V1 = 0.50L

Final Volume, V2 = 0.10L

Initial Temperature, T1 = 15 +273 = 288K (Converting to kelvin temperature)

Final Temperature, T2 = ?

The equation relating all these parameters id the ideal gas equation.

This is given as;

(P1V1) / T1 = (P2V2) / T2

Upon solving for T2, we have;

T2 = P2V2T1 / PIV1

Inserting the values,

T2 = (13 * 0.10 * 288) / (7.6 * 0.5)

T2 = 374.4 / 3.8

T2 = 98.53K or 174.47°C

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Illusion [34]

Answer:

- 130.64°C.

Explanation:

  • We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and P are constant, and have two different values of V and T:

<em>V₁T₂ = V₂T₁</em>

<em></em>

V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.

V₂ = 307.0 L, T₂ = ??? K.

<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>

<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>

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Explanation: I hope this helps!

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