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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.

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Answer:

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