Question

question 4: an engineer designs a rigid steel tank that will be used for fuel delivery. this tank is fitted with a safety valve that opens only if the internal pressure of the tank exceeds 8.00 x 10² torr. the safety valve of the steel tank is tested by filling the steel tank with ch₄ gas at 35°c and 0.991 atm and then placing the tank in boiling water at 100°c. will the tank open and what is the new pressure in torr after the tank is placed in the boiling water? a. the tank will open; 1.20 torr b. the tank will not open; 1.20 torr c. the tank will open; 949 torr d. the tank will not open; 912 torr e. the tank will open; 912 torr

Explanation:

Step1: Convert pressure units

We know that \(1\ atm = 760\ torr\). So the initial pressure \(P_1=0.991\ atm\times760\ torr/atm = 753.16\ torr\)
The initial temperature \(T_1=(35 + 273)K=308\ K\) and the final temperature \(T_2=(100 + 273)K = 373\ K\)

Step2: Apply Gay - Lussac's law

Gay - Lussac's law is \(P_1/T_1=P_2/T_2\). Rearranging for \(P_2\), we get \(P_2=\frac{P_1T_2}{T_1}\)
Substitute \(P_1 = 753.16\ torr\), \(T_1 = 308\ K\) and \(T_2=373\ K\) into the formula:
\(P_2=\frac{753.16\times373}{308}\)
\(P_2=\frac{753.16\times373}{308}\approx912\ torr\)

Since \(912\ torr>8.00\times 10^{2}\ torr\), the tank will open.

Answer:

E. The tank will open; 912 torr