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Chapter 5: Problem 80

Tank A has ammonis at \(300 \mathrm{~K}\), Tank B hes nitrogen gas at \(150 \mathrm{~K}\). Thaks A and B have the same volume. Compare the presrures in tanks \(A\) and \(B\) if (a) tank B has twice as many moles of nitrogen as tank \(A\) has of ammonia. (b) tank A has the same number of moles of ammonia as tank \(\mathrm{B}\) has of nitrogen. (Try to do this without a calculatorl)

Short Answer

Expert verified
Question: Compare the pressures in tank A and tank B under the two given conditions. Answer: For condition (a), the pressure in tank A is half that of tank B. For condition (b), the pressure in tank A is twice that of tank B.

Step by step solution

01

Tank A pressure formula

Using the Ideal Gas Law, \(PV = nRT\), we first note that the Volume (V) and the ideal gas constant (R) are the same for both tanks. So, we can write the pressure and temperature relationship for tank A as: \(P_A = \frac{n_{NH_3}R}{V} * T_{NH_3}\) where \(P_A\) is the pressure in tank A, \(n_{NH_3}\) is the number of moles of ammonia, and \(T_{NH_3} = 300 \mathrm{K}\).
02

Tank B pressure formula

Similarly, for tank B, we can write the relationship between pressure, temperature, and moles of nitrogen as: \(P_B = \frac{n_{N_2}R}{V} * T_{N_2}\) where \(P_B\) is the pressure in tank B, \(n_{N_2}\) is the number of moles of nitrogen, and \(T_{N_2} = 150 \mathrm{K}\). Now let's find the ratios of the pressures in tanks A and B for conditions (a) and (b).
03

Pressure ratio for condition (a)

For condition (a), we are given that the number of moles of nitrogen in tank B is twice that of ammonia in tank A: \(n_{N_2} = 2n_{NH_3}\). Now, let us find the ratio of \(P_A\) to \(P_B\): \(\frac{P_A}{P_B} = \frac{(\frac{n_{NH_3}R}{V} * 300)}{(\frac{(2n_{NH_3})R}{V} * 150)} = \frac{300}{2 * 150} = \frac{1}{2}\) So, in condition (a), the pressure in tank A is half that of tank B.
04

Pressure ratio for condition (b)

Condition (b) states that the number of moles of ammonia in tank A is equal to the number of moles of nitrogen in tank B: \(n_{NH_3} = n_{N_2}\). The ratio of pressures of A and B is: \(\frac{P_A}{P_B} = \frac{(\frac{n_{NH_3}R}{V} * 300)}{(\frac{n_{NH_3}R}{V} * 150)} = \frac{300}{150} = 2\) Hence, for condition (b), the pressure in tank A is twice that of tank B.

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