Question
A sample of argon has a volume of 0.983 L, a pressure of 2.50 atm, and a temperature of 32 °C. What is the new volume, if the new pressure is 1.55 atm and the new temperature is 55 °C?
Answers
GPT 3.5
To solve this problem, we can use the combined gas law equation:
(P1V1)/T1 = (P2V2)/T2
Where:
P1 = 2.50 atm
V1 = 0.983 L
T1 = 32°C + 273 = 305 K
P2 = 1.55 atm
T2 = 55°C + 273 = 328 K
V2 = unknown
(P1V1)/T1 = (P2V2)/T2
(2.50 * 0.983) / 305 = (1.55 * V2) / 328
2.4575 / 305 = 1.55V2 / 328
0.00804 = 0.00473V2
V2 = 0.00804 / 0.00473
V2 = 1.70 L
Therefore, the new volume of the argon sample is 1.70 L.
(P1V1)/T1 = (P2V2)/T2
Where:
P1 = 2.50 atm
V1 = 0.983 L
T1 = 32°C + 273 = 305 K
P2 = 1.55 atm
T2 = 55°C + 273 = 328 K
V2 = unknown
(P1V1)/T1 = (P2V2)/T2
(2.50 * 0.983) / 305 = (1.55 * V2) / 328
2.4575 / 305 = 1.55V2 / 328
0.00804 = 0.00473V2
V2 = 0.00804 / 0.00473
V2 = 1.70 L
Therefore, the new volume of the argon sample is 1.70 L.