Question

If I have 21L of gas held at a pressure

of 78atm and a temperature of 900K,
what will be the temperature of the gas
if I decrease the pressure to 45.2atm and
increase the volume to 30.0L?

Answer 1

T2 = 745 K

Step-by-step explanation:

This is an ideal gas law problem. We can use this formula to find our answer:

(P1*V1)/T1 = (P2*V2)/T2

So we have these given in the problem:

P1 = 78 atm P2 = 45.2 atm

V1 = 21 L V2 = 30.0 L

T1 = 900 K T2 = ?

So we put all of this stuff into the equation and solve for the unknown T2:

[(78 atm)*(21 L)]/(900 K) = [(45.2 atm)*(30.0 L)]/T2

(1638 atm*L)/900 K = (1356 atm*L)/T2

1.82 atm*L/K = (1356 atm*L)/T2

[flip around to put T2 on the numerator and alone]

T2 = (1356 atm*L)/1.82 atm*L/K

[atm*L cancel out to leave us with K]

T2 = 745.0549451

T2 = 745 K