Suppose I am given following reaction in a vessel: $$\ce{NH4HS(s) <=> NH3(g) + H2S(g)}$$ $P_\ce{NH3} = P_\ce{H2S} = \pu{1 atm}$ at equilibrium; if volume is doubled at this instant, then what would be $P_\ce{NH3}$ at new equilibrium?
I know that as soon as volume is doubled, pressure would drop to half at this instant, and $K_\mathrm{c}$ will have the same value at new equilibrium, but how to actually use this to calculate the partial pressure of ammonia at new equilibrium?