A 20.0L container holds 0.650 mol of He gas at 37.0C at a pressure of 628.3torr. The new pressure inside the container will be torr if the volume is reduced to 12.0L, the temperature increased to 177C, and 1.25 mol of additional He gas was added.

2918.4 torr

Explanation:

From;

PV=nRT

We have that the new conditions of the gas are;

V = 12.0 L

T1 = 177 + 273 = 450 K

n = 1.25 mol

R= 0.082 atm LK-1mol-1

P= ?

P = nRT/V

P = 1.25 *  0.082 * 450/12.0

P = 3.84 atm

Since

1 atm = 760 torr

3.84 atm = 3.84 * 760 = 2918.4 torr