The volume of a sample of oxygen is 200.0 mL when the pressure is 3.000 atm and the temperature is 37.0 oC. What is the new temp
erature if the volume increases to 400.0 mL and the pressure decreases to 2.000 atm
1 answer:
Answer:
Explanation:
Given:
V1 = 200 ml
T1 = 20 °C
= 20 + 273
= 293 K
P1 = 3 atm
P2 = 2 atm
V2 = 400 ml
Using ideal gas equation,
P1 × V1/T1 = P2 × V2/T2
T2 = (2 × 400 × 293)/200 × 3
= 234400/600
= 390.67 K
= 390.67 - 273
= 117.67 °C
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