A sample of a monatomic ideal gas is originally at 20 °c. what is the final temperature of the gas if the pressure is doubled and volume is reduced to one-fourth its initial value?

the final temperature is T2 = - 126.5°C

Explanation:

Since the gas behaves as an ideal gas

Initial state) P1*V1=n*R*T1

Final state) P2*V2=n*R*T2

where P = gas pressure, V=volume occupied by the gas, n = number of moles, R = ideal gas constant, T = absolute temperature

dividing both equations

(P2/P1) * (V2/V1) = T2/T1

since P2=2*P1 and V2=1/4*V1, T1=20°C=293K

2*1/4=T2/T1

T2=T1/2 = 293K/2 = 146.5K

T2 = 146.5K = - 126.5°C