Chapter 15: Q57GP (page 412)

(II) When\({\bf{5}}{\bf{.80 \times 1}}{{\bf{0}}{\bf{5}}}\;{\bf{J}}\)of heat is added to a gas enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure, the volume is observed to increase from\({\bf{1}}{\bf{.9}}\;{{\bf{m}}{\bf{3}}}\)to\({\bf{4}}{\bf{.1}}\;{{\bf{m}}{\bf{3}}}\). Calculate
(a) the work done by the gas, and
(b) the change in internal energy of the gas.
(c) Graph this process on a PV diagram.

Short Answer

Expert verified

(a) The work done by the gas is\(2.22 \times {105}\;{\rm{J}}\).

(b) The change in the internal energy of the gas is\(3.58 \times {105}\;{\rm{J}}\)

Step by step solution

Work done at constant pressure

To find the work done at constant pressure, use the product of given pressure and change in volume of the particular gas.

Given data

The heat added is \(Q = 5.80 \times {105}\;{\rm{J}}\).

The initial volume is \({V_1} = 1.9\;{{\rm{m}}3}\).

The final volume is \({V_2} = 4.1\;{{\rm{m}}3}\).

The pressure is \(P = 1\,{\rm{atm}}\).

(a) Evaluation of work done by the gas

The relation for the work done is given by:

\(\begin{array}{l}W = P\Delta V\\W = P\left( {{V_2} - {V_1}} \right)\end{array}\)

Here, \(\Delta V\) is the change in volume.

Substitute the values in the above expression.

\(\begin{array}{l}W = \left( {1\;{\rm{atm}} \times \frac{{1.01 \times {{10}5}\;{\rm{Pa}}}}{{1\;{\rm{atm}}}}} \right)\left( {4.1\;{{\rm{m}}3} - 1.9\;{{\rm{m}}3}} \right)\\W = 2.22 \times {105}\;{\rm{J}}\end{array}\)

Thus, the work done by the gas is\(2.22 \times {105}\;{\rm{J}}\).

(b) Evaluation of the change in internal energy

The relation from the first law of thermodynamics is given by:

\(\Delta U = Q - W\)

Substitute the values in the above expression.

\(\begin{array}{l}\Delta U = \left( {5.80 \times {{10}5}\;{\rm{J}}} \right) - \left( {2.22 \times {{10}5}\;{\rm{J}}} \right)\\\Delta U = 3.58 \times {105}\;{\rm{J}}\end{array}\)

Thus, the change in internal energy of the gas is \(3.58 \times {105}\;{\rm{J}}\).