Answer:
Part a)
![T = 425.6 K](https://img.qammunity.org/2020/formulas/physics/college/jtrde0y5nht5i5yqkqwtc7xi0co3yg2j1n.png)
Part b)
![KE_(avg) = 8.81* 10^(-21) J](https://img.qammunity.org/2020/formulas/physics/college/rz9lof5u2erefgc99bdxkxzzfnws8vaurf.png)
Part c)
in order to find the average speed we need to know about the the gas molar mass or we need to know which gas it is.
Step-by-step explanation:
Part a)
As per ideal gas equation we know that
![PV = nRT](https://img.qammunity.org/2020/formulas/chemistry/middle-school/9s3lu4eymz9b8l00rczismrm9dp9at9je4.png)
here we know that
![P = 1.60 * 10^6 Pa](https://img.qammunity.org/2020/formulas/physics/college/swh3ywlw03wftf7ifw0x6fnesd332ax3ko.png)
n = 3.80 moles
![V = 8.40 L = 8.40 * 10^(-3) m^3](https://img.qammunity.org/2020/formulas/physics/college/21zk6tuyel6axqjmdupe3jgmpg7ry9fbc8.png)
now from above equation we have
![T = (PV)/(nR)](https://img.qammunity.org/2020/formulas/physics/college/u7chlj2vd2wvq37b4z72e7e4278jksn28j.png)
![T = ((1.60 * 10^6)(8.40 * 10^(-3)))/((8.31)(3.80))](https://img.qammunity.org/2020/formulas/physics/college/jixrwmnm86gjf22on443gbowrk08nzd008.png)
![T = 425.6 K](https://img.qammunity.org/2020/formulas/physics/college/jtrde0y5nht5i5yqkqwtc7xi0co3yg2j1n.png)
Part b)
Average kinetic energy of the gas is given as
![KE_(avg) = (3)/(2)KT](https://img.qammunity.org/2020/formulas/physics/college/6bnq91jy8p89z8una71cozbgkdcfre4b89.png)
here we know that
![K = 1.38 * 10^(-23)](https://img.qammunity.org/2020/formulas/physics/college/4ra3xv9qo7x8c74q9eij8tueoa0153sn8n.png)
T = 425.6 K
now we have
![KE_(avg) = (3)/(2)(1.38 * 10^(-23))(425.6)](https://img.qammunity.org/2020/formulas/physics/college/c6xom99b0merctbgn6r6kmv0lkiuf7e1oj.png)
![KE_(avg) = 8.81* 10^(-21) J](https://img.qammunity.org/2020/formulas/physics/college/rz9lof5u2erefgc99bdxkxzzfnws8vaurf.png)
Part c)
in order to find the average speed we need to know about the the gas molar mass or we need to know which gas it is.