A given mass of gas at a temperature of 30°C is trapped in a tube of volume V. Calculate the temperature of the gas when the volume is reduced to two-third of its original value by applying a pressure twice the original value.
71oC
40oC
131oC
313oC
404oC
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Discussions (16)
My school is correct since the options provided are in °C which can as well be 404K equivalent.
Explanation given below.
I pray you understand as you go through it. Amen 
V=kT; when T is reduced to 2/3 of volume,
T becomes 2/3(273+30)=202K
P=kT; when P is doubled T is also doubled; Hence T becomes 2×202=404k
But K=°C+273
404=°C+273
°C=404-273
= 131°C
Thanks for your contributions. If you feel the answer is not correct, please give a detailed explanation to back up your answers.
The answer is 404Kelvin Or assuming The 30• reps change in Temperature the answer can then either be 53.3°C or K
Here is an explanation:
(P1*V1)/T1=(P2*V2)/T2
Make T2 the subject of the formula, that is,
(P2*V2*T1)/(P1*V1)=T2
P1=P1 P2=2P1
T1=30 C=303K T2=?
V1=V V2=2V1/3
(2*P1*0.67*V1*303)/(P1*V1)
YOU ARE LEFT WITH 2*0.67*303=406.02K=133.02C
I'll clarify
The Ans is C cuz 4/3 × 303K (has to be done in kelvin) = 404K
404K= 131°C... hence the Ans is C
P1V1/T1=P2V2/T2......P1V1/303=2P2×2/3V2/T2
When the p and v in both side of the equation cancels...we hv ....T2=303×2×2/3.....T2=1212/3=404....so the answer is supposed to be D and not C
The correct answer is C...the celsius must be converted to kelvin before solving it.
