Explanation Given data: v rms is 1223 m s . m is 3.32 × 10 − 26 kg . Mass and rms speed of each tank is shown below. A B C D Mass m m 2 m 2 m Rms speed v rms 2 v rms v rms 2 v rms Formula used: Consider the expression for kinetic theory of gas. 1 2 m v rms 2 = 3 2 k T (1) Here, m is the mass, v rms is the root mean squared speed, k is the Boltzmann’s constant, which is known as 1.38 × 10 − 23 J K , and T is the temperature. Explanation: Rearrange equation (1). T = m v rms 2 3 k (2) Case A: Modify equation (2). T A = m A v rms,A 2 3 k Substitute 1.38 × 10 − 23 J K for k , 3.32 × 10 − 26 kg for m A and 1223 m s for v rms,A , T A = ( 3.32 × 10 − 26 kg ) ( 1223 m s ) 2 3 ( 1.38 × 10 − 23 J K ) = ( 3.32 × 10 − 26 ) ( 1.495 × 10 6 ) 4.14 × 10 − 23 K = 4.9634 × 10 − 20 4.14 × 10 − 23 K = 1198.88 K T A ≅ 1200 K Case B: Modify equation (2). T B = m B ( 2 v rms,B 2 ) 3 k Substitute 1.38 × 10 − 23 J K for k , 3.32 × 10 − 26 kg for m B and 1223 m s for v rms,B , T B = ( 3.32 × 10 − 26 kg ) ( 2 ( 1223 m s ) ) 2 3 ( 1.38 × 10 − 23 J K ) = ( 3.32 × 10 − 26 kg ) ( 2446 m s ) 2 3 ( 1.38 × 10 − 23 J K ) = ( 3.32 × 10 − 26 ) ( 5.982 × 10 6 ) 4

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Chapter 14, Problem 34P

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The temperature of the gas in four tanks A, B, C and D.

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Chapter 14, Problem 33P

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