Explanation
Step 1: Preparing data
Step 2: Solving using equations
Correct Option: $$B$$
Hint
Change in enthalpy, $$\Delta H=n{{C}_{P}}\Delta T$$
Enthalpy change will be positive as temperature changes can never be negative.
And both the constant term is positive,
Thus,
$$\Delta H=\left( + \right)ve$$
When the temperature fluctuates on this graph, the enthalpy changes as well.
Again, from the gas equation it is clear that,
$$~V\propto T$$
When the temperature rises, the volume of the gas rises with it.
$$W=-P\left( {{V}_{2}}-{{V}_{1}} \right)$$
As volume increases so $$\left( {{V}_{2}}-{{V}_{1}} \right)$$ is positive.
So, amount of work done is positive.
Final answer
The correct answer is option (A).
$$2SO_2 + O_2 \rightarrow2SO_3$$
$$K_p$$ can be calculated as:
$$K_p=\dfrac{[P_{SO_3^{2-}}]}{[P_{SO_2}^2]\times [P_{O_2}]}$$
As given:
$$[P_{SO_3}]=[P_{SO_2}]$$
So,
$$5=\dfrac{1}{[O_2]}$$
$$P_{O_2}=\dfrac{1}{5}$$
$$P_{O_2}=0.2$$
So, partial pressure of oxygen will be 0.2 atm.
0.2 atm will be the partial pressure of $$O_2$$.
Please disable the adBlock and continue. Thank you.