Question

4.A 1.00-mol sample of ammonia at 14.0 atm and 25°C in a cylinder fitted with a movable piston expands against a constant external pressure of 1.00 atm. At equilibrium, the pressure and volume of the gas are 1.00 atm and 23.5 L, respectively. (a) Calculate the final temperature of the sample. (b) Calculate the values of q, w, and dU for the process ​

Answer 1

Explanation:

Assume ammonia is ideal gas. the final temp will be

$T_{2} = \frac{P_{2} V_{2} }{nR}$

= $\frac{(1 atm)(23.5L)}{(1 mol)(0.08206)}$

= 286K

The final vol is given but intial vol needs to be calculated from ideal gas law

$V_{1} = \frac{nRT_{1} }{P_{1} }$

= $\frac{(1)(0.08206)(298)}{14}$

= 1.75L

work done is

W = (1)(23.5L - 1.75L)$\frac{101.3}{1}$

= -2.20 x $10^{3}$ J

It is assumed that Cv and Cp ae independent

Cv= Cp - nR

=27.35 J/K

Therfore, internal energy is

U = CvT

= 27.5 x (286K - 298K)

= -330 J

Using first law

q = -330 + 2200

q = 1870J

Answer 2

GIVEN:

          Atmospheric pressure of ammonia is($P_{1}$) $14 atm$

          N.o of moles is $(N)$ $1mole$

          Temperature of ammonia $(T{1} )$ is $25$°$C$ =$298K$

           Pressure of the gas $(P_{2} )= 1atm$

           volume of the gas is $(23.5 l)$

          THE SOLUTION is in the below picture .......