One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium.
A(g) $$\rightleftharpoons$$ B(g)
K$$_c$$ for this reaction at 400 K is 4.0. The partial pressures (in atm) of He and B(g) are respectively (at equilibrium)
(Assume He, A(g) and B(g) behave as ideal gases)
(Given: R = 0.082 L atm K$$^{-1}$$ mol$$^{-1}$$)

For the equilibrium

$$A(g)\rightleftharpoons B(g),$$

let (x) moles of (A) be converted into (B).

At equilibrium,

$$n_A=1-x,\qquad n_B=x,\qquad n_{He}=1.$$

Using the expression for the equilibrium constant,

$$K_c=\frac{[B]}{[A]}=\frac{x/V}{(1-x)/V}=\frac{x}{1-x}=4.$$

Solving,

$$x=4(1-x),$$

$$x=4-4x,$$

$$5x=4,$$

$$x=0.8.$$

Therefore, the equilibrium amounts are

$$n_A=0.2\ \text{mol},\qquad n_B=0.8\ \text{mol},\qquad n_{He}=1.0\ \text{mol}.$$

Using the ideal gas equation,

$$P=\frac{nRT}{V},$$

with

$$R=0.082\ \text{L atm K}^{-1}\text{mol}^{-1},\qquad T=400\ \text{K},\qquad V=10\ \text{L},$$

we obtain

$$\frac{RT}{V}=\frac{0.082\times400}{10}=3.28\ \text{atm mol}^{-1}.$$

Hence, the partial pressure of helium is

$$P_{He}=1.0\times3.28=3.28\ \text{atm}.$$

The partial pressure of (B(g)) is

$$P_B=0.8\times3.28=2.624\ \text{atm}.$$

Therefore, the equilibrium partial pressures are

$$P_{He}=3.28\ \text{atm},\qquad P_B=2.624\ \text{atm},$$

which corresponds to Option A.