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

Answer: 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