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**Hint**: Pressure -Volume work done occurs when the volume of the gas or system changes. Work done by a system is the energy transferred from it to its surroundings. Express that work in joules and calories, -${{P}_{external}}$$\times $$\vartriangle V$.

**Complete step by step answer:**

If the gas is heated, the gas molecules are energized. The increase in average kinetic energy of the gas molecules can be observed by measuring how the temperature of the gas particles increased. The gas molecules collide with the piston more frequently, by the fast movement of the gas particles. These frequent collisions transfer energy to the piston and allow it to move against an external pressure thus, increasing the net volume of the gas.

The work done by the gas is specifically the expansion and compression of the gas. Work done by gases is sometimes called pressure-volume

__or__

__PV__work.

To calculate how much work is done (or has done to it) by the gas against a constant external pressure, we have to use an equation:

WORK= (w) = -${{P}_{external}}$$\times $$\vartriangle V$ where external pressure is denoted by ${{P}_{external}}$. The external pressure opposes the pressure of the gas in the system. ∆V is a change in the volume of the gas, which is calculated as ${{V}_{final}}-{{V}_{\operatorname{in}itial}}$.

Sign convention of work :

When a system does work on the surroundings, negative work is said to have occurred.

When the gas does work, the volume of the gas particles decreases ($\vartriangle V$<0).The work done is positive.

When the gas does work, the volume of the gas particles increases ($\vartriangle V$>0).The work done is negative.

Units conversion:

Work has its units in JOULES.

Pressure-volume work has its unit in L-atm which is to be converted to Joules using the conversion factor 101.325J/1L-atm

To convert joules to calories the conversion factor is 0.239cal/1J.

Calculate the pressure-volume work done by the gas:

Given: ${{P}_{external}}$= 10 atmospheres ${{V}_{initial}}$= 1 litre ${{V}_{final}}$ = 2 litres

Use the formula, to calculate the work done by the gas:

WORK= (w) = -${{P}_{external}}$$\times $$\vartriangle V$

Substitute the values as ${{V}_{final}}-{{V}_{\operatorname{in}itial}}$

The net formula is

= -${{P}_{external}}$$\times $(${{V}_{final}}-{{V}_{\operatorname{in}itial}}$)

Substitute the values of ${{P}_{external}}$,${{V}_{initial}}$, ${{V}_{final}}$ as 10 atmospheres, 1 litre and 2 litres.

The work done is

= - (10 atm)$\times $(2 litres – 1 litre)

= - (10 atm)$\times $(1 litre)

= - (10 litre-atm)

Convert litre atm to joules using the conversion factor:

1 L-atm has 101.325J, so -10 L-atm has (-10)$\times $(101.325J) is equal to (-1013.25 Joules. )

To convert joules into calories multiply the joule by 0.239 factor, (0.239) $\times $(-1013.25) is equal to - 242.16 Calories.

The answer of the question is -1013.25 Joules and -242.16 calories.

**Note**:

(i) Take care of the sign convention while solving the numerical of the work done. As the sign convention by the gas and on the gas is different.

(ii) Do not forget to convert to use conversion factor for the required units as mentioned in the question.

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