What happens to the pressure of a sample of helium gas if the temperature is increased from 200 K to 800 K with no change volume?
Correct answer: A. Pressure increases by a factor of 4
- A. Pressure increases by a factor of 4
- B. Pressure decreases by a factor of 4
- C. Pressure decreases by a factor of 2
- D. Pressure increases by a factor of 2
- E. No change in pressure
Explanation
According to combined gas law, P1V1/T1 = P2V2/T2, The volume is constant so we can say that pressure and temperature are directly propotional. From 200K to 800K, the temperature increases four times (200x4=800) and since the pressure and temperature are directly propotional to eachother, the pressure also increases by a factor of 4.We all know ideal gas equation,Easy way to solve such questions PV = nRTHere V is given constant ( no change)for a sample n is also constantR is general gas constant So pressure is directly proportional to temperature So ratio will beP1/ P2 = T1/ T2We have to find P2 as they said after change in temperatureT1 = 200 KT2= 800 KPut,P1/ P2 = 200 / 800P1/ P2 = ¼Cross multiplying 4P1 = P2As we see the final pressure is 4 times the initial pressure , So ,Pressure increases by a factor of 4There's another method to solve We know ,T1 = 200 KT2= 800 KWe can write T2 asT2 = 4 × 200 But 200 K is T1 So we can write T2 = 4 × T1 Means that the final temperature is increased 4 times the initial temperature.But we also know Pressure is directly proportional to temperature If temperature is increases by 4 times, obviously pressure will also be increased by 4 times
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