Question

A balloon containing an ideal gas has a volume of 10 litre and temperature of $17^oC$ . If it is heated slowly to $75^0C$, the work done by the gas inside the balloon is $2 \times 10^x$ J. Find x .(neglect elasticity of the balloon and take atmospheric pressure as $10^5$ Pa)

Answer 1

Hint
Thus, the heat given off or absorbed during a chemical reaction at constant pressure is equal to the change in the enthalpy of the system. The relationship between the change in the internal energy of the system during a chemical reaction and the enthalpy of reaction can be summarized as follows.

Complete step by step answer
According to the question the balloon is heated slowly to $17^0C$ to $75^0C$
So, the initial temperature $T_{ini} = 17^0C$
And, the final temperature $T_{final} = 75^0C$
And, we know Patm = $10^5$ pa
So, know work done $W = {P_{atm}}\Delta V$ ...............equation 1
Now we can also say,
$\Rightarrow W = {P_{atm}}({V_{final}} - {V_{ini}})$
We know that $\dfrac{{{V_{ini}}}}{{{T_{ini}}}} = \dfrac{{{V_{final}}}}{{{T_{final}}}}$
Now , continuing the equation we get that,
$\Rightarrow W = {P_{atm}}(\dfrac{{{V_{ini}}{T_{final}}}}{{{T_{ini}}}} - {V_{ini}})$
$\Rightarrow W = {P_{atm}}{V_{ini}}(\dfrac{{{T_{final}}}}{{{T_{ini}}}} - 1) $
Now putting the values that we know,
We get, $W = {10^5} \times 10 \times {10^{ - 3}}(\dfrac{{273 + 75}}{{273 + 17}} - 1)$ [here changing the liter into $m^3$]
So, the work done $W = 200J = 2 \times {10^2}J$
Comparing the value with the given value we get, X = 2.

Note
Demand is said to be elastic if the change in price causes a more than proportionate change in quantity demanded. At 10 p.c. a change in price causes quantity demanded to change by more than 10 p.c. In other words, if E is greater than one, demand is said to be elastic