A given mass of gas at a temperature of 30^\circC 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.
A
71oC
B
40oC
C
131oC
D
313oC
E
404oC
Verified Correct Answer: C.
Physical Principles & Numerical Solution
Using the genaral gas law: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)
T\(_1\) = 30º = 273 + 30 = 303k
T\(_2\) = ?
let V\(_1\) = V, V\(_2\) = \(\frac{2}{3}\)V, P\(_1\) = P, P\(_2\) = 2P
From, \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)
T\(_2\) = \(\frac{P_2 V_2 T_1}{P_1 V_1}\) = \(\frac{2P \times \frac{2}{3}V \times 303}{P \times V}\) = 404k(the Ps and Vs cancel out)
404k → ºC = 404 - 273 = 131ºC
Source: Myschool Past Questions Archive (Record #24256)