What is the pressure of a 59.6-L gas sample containing 3.01 mol of gas at 44.9°C? (R = 0.0821 L • atm/(K • mol),1 atm = 760 mmHg) A) 1.41 $\times$ 10² mmHg B) 1.73 $\times$ 10^-3 mmHg C) 1.32 mmHg D) 1.00 $\times$ 10³ mmHg E) 5.77 $\times$ 10² mmHg

Question

Quizplus · Accepted Answer

We can use the ideal gas law to solve for the pressure:
PV = nRT
P = nRT/V
P = (3.01 mol)(0.0821 L•atm/(K•mol))(44.9°C+273.15) / 59.6 L
P = 1.00×10³ mmHg (using the conversion factor 1 atm = 760 mmHg)

Therefore, the answer is D) 1.00×10³ mmHg.

What Is the Pressure of a 59 ×\times× 102 MmHg B) 1

What Is the Pressure of a 59 $\times$ 10² MmHg
B) 1