What is the final pressure of 1.00 mol of ammonia gas, initially at 1.00 atm, if the volume is:. a. gradually decreased from 78.0 mL to 39.0 mL at constant temperature.
1) 2.00 atm
2) 1.00 atm
3) 0.50 atm
4) Not enough data
b. increased from 43.5 mL to 65.5 mL at constant temperature.
c. decreased by 40% at constant temperature.

The correct answer is a = 2 atm, b = 0.66 atm, and c = 1.66 atm.

Explanation:

a. When the temperature is constant, the ideal gas equation, that is, PV = nRT becomes PV = constant, or P1V1 = P2V2

Based on the given information, the pressure of the gas is 1.00 atm, the V1 of the gas is 78 ml, and the V2 of the gas is 39 ml.  

Now using the above formula we get,  

P2 = (P1V1) / (V2)

P2 = 1 atm * 78 ml/39 ml

P2 = 2 atm.  

b) Now the volume V1 is 43.5 ml, and the volume V2 is 65.5 ml at constant temperature, P1 is 1 atm. Now P2 will be,  

P2 = (P1V1) / (V2)

P2 = 1 atm * 43.5 ml/65.5 ml

P2 = 0.66 atm

c) In the given case, when the volume is decreased by 40 percent, let us consider that the volume, V1 be 100 ml, and the volume V2 be 60 ml, the P1 is 1 atm, now the P2 will be,  

P2 = (P1V1) / (V2)

P2 = 1 atm * 100 ml/60 ml

P2 = 1.66 atm.

the correct answer is:

a. formation of precipitation

the answer is c

tell me if wrong