A 35.8 L cylinder of \(\mathrm{Ar}(\mathrm{g})\) is connected to an evacuated 1875 L tank. If the temperature is held constant and the final pressure is \(721 \mathrm{mmHg}\), what must have been the original gas pressure in the cylinder, in atmospheres?

Short Answer

Expert verified
The original gas pressure in the cylinder was 50.71 atmospheres.

Step by step solution

01

Identify known and unknown variables

The problem provides the following: Initial volume \(V_1\) = 35.8 L, Final volume \(V_2\) = 35.8 + 1875 = 1910.8 L, Final pressure \(P_2\) = 721 mmHg, Temperature \(T\) is constant and Initial pressure \(P_1\) is unknown. The goal is to find \(P_1\).
02

Convert pressure to correct units

The pressure in the problem is given in mmHg, but the standard unit in the ideal gas law is atmospheres. Thus, convert final pressure from mmHg to atmospheres: \(P_2 = 721 mmHg * \frac{1 atm}{760 mmHg} = 0.95 atm.\)
03

Apply Boyle's law

Boyle's law is essentially a restatement of the ideal gas law and states that the pressure and volume of a gas have an inverse relationship when temperature is held constant. Hence, we can write its formula as \(P_1V_1 = P_2V_2\). Substitute the known values into the equation and solve for \(P_1\): \(P_1 = \frac{P_2V_2}{V_1} = \frac{0.95 atm * 1910.8 L}{35.8 L}= 50.71 atm\).

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Most popular questions from this chapter

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