Q4P (II) A gas is enclosed in a cyli... [FREE SOLUTION]

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Chapter 15: Q4P (page 412)

(II) A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 254 kcal of heat is added to the gas, the volume is observed to increase slowly from to \({\bf{16}}{\bf{.2}}\;{{\bf{m}}^{\bf{3}}}\).Calculate (a) the work done by the gas and (b) the change in internal energy of the gas.

Short Answer

Expert verified

The work done is \(4.2 \times {10^5}\;{\rm{J}}\).
The change in internal energy is \(6.4 \times {10^5}\;{\rm{J}}\).

Step by step solution

Concepts

From the first law of thermodynamics, \(\Delta U = Q - W\).

The work done is\(W = P\Delta V\).

Given data

The pressure is \(P = 1\;{\rm{atm}} = 1.01 \times {10^5}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

The initial volume of the gas is \({V_1} = 12.0\;{{\rm{m}}^{\rm{3}}}\).

The final volume is the gas is \({V_2} = 16.2\;{{\rm{m}}^{\rm{3}}}\).

The heat added is

\(\begin{array}{c}Q = 254\;{\rm{kcal}}\\ = 254 \times {10^3}\;{\rm{cal}}\\ = 254 \times {10^3} \times 4.186\;{\rm{J}}{\rm{.}}\end{array}\)

Calculation

Part (a)

The work done is calculated below:

\(\begin{array}{c}W = P\Delta V\\ = P\left( {{V_2} - {V_1}} \right)\\ = \left( {1.01 \times {{10}^5}\;{\rm{N/}}{{\rm{m}}^2}} \right) \times \left( {16.2\;{{\rm{m}}^{\rm{3}}} - 12.0\;{{\rm{m}}^{\rm{3}}}} \right)\\ = 4.2 \times {10^5}\;{\rm{J}}\end{array}\)

Hence, the work done is \(4.2 \times {10^5}\;{\rm{J}}\).

Part (b)

From the first law of thermodynamics, you get

\(\begin{array}{c}\Delta U = Q - W\\\Delta U = \left( {254 \times {{10}^3} \times 4.186\;{\rm{J}}} \right) - \left( {4.2 \times {{10}^5}\;{\rm{J}}} \right)\\\Delta U = 6.4 \times {10^5}\;{\rm{J}}{\rm{.}}\end{array}\)

Hence, the change in internal energy is \(6.4 \times {10^5}\;{\rm{J}}\).