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A frictionless piston cylinder device and a rigid tank contains 5 kg of helium gas at the same...

Question:

A frictionless piston cylinder device and a rigid tank contains 5 kg of helium gas at the same temperature, pressure and volume. Now heat is transformed and the temperature of both system is dropped by 12{eq}^o {/eq}C. The amount of extra heat that most be supplied to helium gas in the cylinder that is maintained as constant pressure is:

a) 125 kJ,

b) 0 kJ,

c) 187 kJ,

c) 499 kJ.

Heat capacity

The content of heat energy that is sufficient to increase the temperature of a body of unit mass by unit degree Celsius, is called the heat capacity of that body.

Answer and Explanation:

Given data:

  • The mass of the helium gas is, {eq}m = 5\;{\rm{kg}} {/eq}
  • Drop in temperature of both the systems is, {eq}\Delta T = 12\;^\circ {\rm{C}}\left( {{\rm{or}}\;{\rm{12}}\;{\rm{K}}} \right) {/eq}

Write the expression for the characteristic gas constant of the helium gas.

{eq}R = {C_p} - {C_v} {/eq}

Here, the specific heat capacity at constant volume is {eq}{C_v} {/eq} and the specific heat capacity at constant pressure is {eq}{C_p} {/eq}

The value of {eq}C_p {/eq}

for helium is {eq}5.1926\;\rm{kJ/kg\cdot{\rm{K}}} {/eq}and the value of {eq}{C_v} {/eq} for helium is {eq}3.1156\;\rm{kJ/kg\cdot{\rm{K}}} {/eq}.

Substitute the values in the above expression.

{eq}\begin{align*} R&=5.1926\;\rm{kJ/kg\cdot{\rm{K}}}-3.1156\;\rm{kJ/kg\cdot{\rm{K}}}\\ &=2.077\;\rm{kJ/kg\cdot{\rm{K}}} \end{align*} {/eq}

Write the expression for the heat supplied to maintain constant pressure.

{eq}Q = mR\Delta T {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} Q&=5\;\rm{kg}\times2.077\;\rm{kJ/kg\cdot{\rm{K}}}\times{12\;\rm{K}}\\ &=124.62\;\rm{kJ}\\ &\approx125\;\rm{kJ} \end{align*} {/eq}

Therefore, {eq}125\;{\rm{kJ}} {/eq} of extra heat must be supplied to the helium gas.


Learn more about this topic:

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The Laws of Thermodynamics

from CLEP Biology: Study Guide & Test Prep

Chapter 2 / Lesson 11
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