#
**Problem**: The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50°C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at 50°C has a total translational kinetic energy of 4000 J.(A) Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen. The root-mean-square speed vrms for diatomic oxygen at 50°C is:(B) The total translational kinetic energy of 1.0 mole of diatomic oxygen at 50°C is:(C) The temperature of the diatomic hydrogen gas sample is increased to 100°C. The root-mean-square speed vrms for diatomic hydrogen at 100°C is: i. (16)(4000 J) = 64000 J ii. (4)(4000 J) = 16000 Jiii. 4000 Jiv. (1/4)(4000 J) = 1000 Jv. (1/16)(4000 J) = 150 Jvi. (√2)(2000 m/s) = 2800 m/s vii. (2)(2000 m/s) = 4000 m/s viii. (1/√2) (2000 m/s) = 1400 m/six. (1/2)(2000 m/s) = 1000 m/s

###### FREE Expert Solution

We are given for a diatomic Hydrogen, H_{2}: its root-mean-square speed at 50°C is 2000 m/s, and the translational kinetic energy of 1 mole at 50°C is 4000 J.

**Part A: **

We are asked to calculate the root-mean-square speed (v_{rms}) of **diatomic oxygen** which has a molar mass that is 16 times that of the diatomic hydrogen at 50°C.

Recall that the **root-mean-square speed** is the measure of the speed of particles in a gas, defined as the *square root of the average velocity-squared of the molecules in gas*.

This is calculated as follows:

Where:

*μ _{rms }*

*= rms speed*

**R **= 8.314 J/mol∙K

**T **= Temperature, Kelvin

**M*** = molar mass, kg/mol*

Given:

*T =** 50°C*

*(Convert to Kelvin)*

*T = 50 + 273.15*

*T = **323.15 K*

###### Problem Details

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50°C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at 50°C has a total translational kinetic energy of 4000 J.

(A) Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen. The root-mean-square speed v_{rms} for diatomic oxygen at 50°C is:

(B) The total translational kinetic energy of 1.0 mole of diatomic oxygen at 50°C is:

(C) The temperature of the diatomic hydrogen gas sample is increased to 100°C. The root-mean-square speed v_{rms} for diatomic hydrogen at 100°C is:

i. (16)(4000 J) = 64000 J

ii. (4)(4000 J) = 16000 J

iii. 4000 J

iv. (1/4)(4000 J) = 1000 J

v. (1/16)(4000 J) = 150 J

vi. (√2)(2000 m/s) = 2800 m/s

vii. (2)(2000 m/s) = 4000 m/s

viii. (1/√2) (2000 m/s) = 1400 m/s

ix. (1/2)(2000 m/s) = 1000 m/s

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Kinetic Energy of Gases concept. You can view video lessons to learn Kinetic Energy of Gases Or if you need more Kinetic Energy of Gases practice, you can also practice Kinetic Energy of Gases practice problems .

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Cheng's class at UCR.