Question

A frictionless piston-cylinder based enclosure contains some amount of gas at a pressure of $400\,{\text{kPa}}$. Then heat is transferred to the gas at constant pressure in a quasi-static process. The piston moves up slowly through a height of $10\,{\text{cm}}$. If the piston has a cross-section area of $0.3\,{{\text{m}}^2}$, the work done by the gas in this process is
A. $6\,{\text{kJ}}$
B. $12\,{\text{kJ}}$
C. $7.5\,{\text{kJ}}$
D. $24\,{\text{kJ}}$

Answer 1

Use the formula for volume of an object in terms of its area and height. Also use the formula for pressure volume work done by the gas at constant pressure. This formula gives the relation between the pressure of the gas and change in volume of the gas. Substitute the values of all the quantities in this formula and calculate the work done by the gas.

Formulae used:
The pressure volume work done $W$ by the gas at constant pressure is given by
$W = P\Delta V$ …… (1)
Here, $P$ is the pressure of gas and $\Delta V$ is the change in volume of the gas.

The volume $V$ of an object is given by
$V = Ah$ …… (2)
Here, $A$ is the area of the object and $h$ is the height of the object.

Complete step by step answer:
We have given that the pressure of the gas is $400\,{\text{kPa}}$.
$P = 400\,{\text{kPa}}$

The height by which the piston moves up is $10\,{\text{cm}}$.
$\Delta h = 10\,{\text{cm}}$

We have also given that the cross-sectional area of the piston is $0.3\,{{\text{m}}^2}$.
$A = 0.3\,{{\text{m}}^2}$

Convert the unit of pressure from kilopascal to Pascal.
$P = \left( {400\,{\text{kPa}}} \right)\left( {\dfrac{{{{10}^3}}}{{1\,{\text{k}}}}} \right)$
$\Rightarrow P = 4 \times {10^5}\,{\text{Pa}}$

Convert the unit of height to SI system of units.

2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)$$ $$ \Rightarrow \Delta h = 0.1\,{\text{m}}$$ We can determine the work done by the gas using equation (1). The change in the volume of the gas is given by equation (2). $$\Delta V = A\Delta h$$ Substitute $$A\Delta h$$ for $$\Delta V$$ in equation (1). $$W = P\left( {A\Delta h} \right)$$ $$ \Rightarrow W = PA\Delta h$$ Substitute $$4 \times {10^5}\,{\text{Pa}}$$ for $$P$$, $$0.3\,{{\text{m}}^2}$$ for $$A$$ and $$0.1\,{\text{m}}$$ for $$\Delta h$$ in the above equation. $$ \Rightarrow W = \left( {4 \times {{10}^5}\,{\text{Pa}}} \right)\left( {0.3\,{{\text{m}}^2}} \right)\left( {0.1\,{\text{m}}} \right)$$ $$ \Rightarrow W = 12000\,{\text{J}}$$ $$ \Rightarrow W = 12\,{\text{kJ}}$$ Therefore, the work done by the gas is $$12\,{\text{kJ}}$$. **Hence, the correct option is B.** **Note:** The students should not forget to convert the unit of pressure of gas from kilopascal to Pascal. Also the students should not forget to convert the unit of change in height of the piston from CGS system of units to the SI system of units. If these unit conversions are not done then we will not obtain the final correct answer.