Two moles of an ideal monatomic gas are at a temperature of 345 k. then, 2250 j of heat is added to the gas, and 870 j of work is done on it. what is the final temperature of the gas

The final temperature is 470K

Explanation: Using the relation;

Q= ΔU +W

Given, n = 2mol

Initial temperature T1= 345K

Heat =Q= 2250J

Workdone=W=-870J(work is done on gas)

T2 =Final temperature =?

ΔU =3/2nR(T2-T1)

ΔU=3/2 × 2 ×8.314 (T2 - 345)

ΔU=24.942(T2-345)

Therefore Q = 24.942(T2-345)+ (-870)

2250=24.942(T2-345)+ (-870)

125.09=(T2-345)

T2 =470K

Therfore the final temperature is 470K

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr