Question

Question: An ideal monatomic gas is contained in a tall cylindrical jar of cross-sectional area \({\bf{0}}{\bf{.080}}\;{{\bf{m}}^{\bf{2}}}\) fitted with an airtight frictionless 0.15-kg movable piston. When the gas is heated (at constant pressure) from 25°C to 55°C, the piston rises 1.0 cm. How much heat was required for this process? Assume atmospheric pressure outside. (Hint: See Section 14–2.)

Short answer

The heat required for this process is \(202\;{\rm{J}}\).

Step-by-step solution

Determination of pressure

The pressure can be obtained by dividing the force exerted on an object by the cross-sectional area of the object. Its value is inversely related to the value of the cross-sectional area of the object.

Given information

The cross-sectional area of the cylinder is \(A = 0.080\;{{\rm{m}}^{\rm{2}}}\).

The atmospheric pressure or outside pressure is \({P_{\rm{o}}} = 1.01 \times {10^5}\;{\rm{Pa}}\).

The mass of the piston is \(m = 0.15\;{\rm{kg}}\).

The displacement of the piston is \(\Delta y = 1.0\;{\rm{cm}}\).

Evaluation of the pressure inside the cylindrical jar

The pressure inside the cylindrical jar is calculated below:

\(\begin{aligned}{c}{P_{\rm{i}}} &= {P_{\rm{o}}} + \frac{{mg}}{A}\\{P_{\rm{i}}} &= \left( {1.01 \times {{10}^5}\;{\rm{Pa}}} \right) + \frac{{\left( {0.15\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)}}{{\left( {0.080\;{{\rm{m}}^{\rm{2}}}} \right)}}\\{P_{\rm{i}}} &= 1.0102 \times {10^5}\;{\rm{Pa}} \approx {\rm{1}}{\rm{.01}} \times {\rm{1}}{{\rm{0}}^5}\;{\rm{Pa}}\end{aligned}\)

The change in the pressure is negligible; therefore, the system can be treated as a constant pressure process.

Evaluation of the heat required for the process

The heat required for the process is calculated below:

\(\begin{aligned}{l}Q &= \Delta U + W\\Q &= \frac{3}{2}P\Delta V + P\Delta V\\Q &= \frac{5}{2}P\Delta V\\Q &= \frac{5}{2}PA\Delta y\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}Q &= \frac{5}{2}\left( {1.01 \times {{10}^5}\;{\rm{Pa}}} \right)\left( {0.080\;{{\rm{m}}^{\rm{2}}}} \right)\left( {1.0\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)\\Q &= 202\;{\rm{J}}\end{aligned}\)

Thus, the heat required for this process is \(202\;{\rm{J}}\).