Question

Suppose that 10 moles of an ideal gas have a gauge pressure of 2 atm and a temperature of 200 K. If the volume of the gas is doubled and the pressure dropped to a gauge pressure of 1 atm, what is the new temperature?
Select one:

a.

267 K

b.

300 K

c.

400 K

d.

200 K

Answer 1

To solve this problem we will apply the concepts related to the equations of ideal gas, from there, we will define the proportion of these values between two states of matter. The ideal gas equation is defined as

$PV = nRT$

Here,

P = Pressure

V = Volume

n = Amount of mass

R = Gas ideal constant

T = Temperature

Between two states we have,

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

The pressure at each state is,

$P_1 = P_{atm}+2 = 3atm$

$P_2 = P_{atm}+1 = 2atm$

According to the statement,

$V_2 = 2V_1$

Then replacing,

$T_2(3atm)(V_1) = T_1(2atm)(2V_1)$

$T_2 = \frac{4}{3}T_1$

$T_2 = \frac{4}{3} 200$

$T_2 = 267K$

Therefore the correct answer is A.