For a certain gas in a closed container, the pressure has been raised by #0.4%#, and the temperature was raised by #"1 K"#. What temperature did the gas start at?

#a)# #"250 K"#
#b)# #"200 K"#
#c)# #"298 K"#
#d)# #"300 K"#

Question

#a)# #"250 K"#
#b)# #"200 K"#
#c)# #"298 K"#
#d)# #"300 K"#

Mateo Aguilar · Accepted Answer

This is an impossible question. However, if we assume that the question should have written that the vessel is rigid, then I get an initial temperature of #"250 K"#.

As is customary, we assume ideality when we see the terms pressure, temperature, and "closed vessel" together.

#PV = nRT#

The closed vessel tells us nothing more than that the mols of gas inside it are constant; it makes no indication of the vessel's volume being constant, since the closed vessel might just be a big, fat balloon.

It seems that we are given:

#P -> 1.004P#

#T -> T + 1#

#n -> n#

#V -> ??? xx V#

Replace to obtain:

#1.004P cdot V = nR(T+1) = nRT + nR#,

where the written variables are all for the initial state and we interpret the question to mean a closed AND rigid vessel. So, we assume that #??? = 1#.

Note that since #PV = nRT#, we can now write:

#1.004nRT = nR(T + 1)#

Then, divide by #nR# to get:

#1.004T = T + 1#

#=> 0.004T = 1#

#=> color(blue)(T = 1/0.004 = "250 K")#

which is one of the options for the given answer; this does not preclude the question from being revised; in fact, it is most likely the intended meaning of the question.

Cameron Andrews · Answer

undefined To find the initial temperature of the gas, you can use the ideal gas law:

$$PV = nRT$$

Where:
- $P$ is the pressure of the gas
- $V$ is the volume of the gas (constant since the container is closed)
- $n$ is the number of moles of gas (constant since the container is closed)
- $R$ is the ideal gas constant
- $T$ is the temperature of the gas

Since the pressure and temperature have changed, we can write:

$$\frac{{P_2}}{{P_1}} = \frac{{T_2}}{{T_1}}$$

Given:
- $\frac{{P_2}}{{P_1}} = 1 + 0.4\% = 1.004$
- $\Delta T = T_2 - T_1 = 1 \, K$

Substituting the values:

$$1.004 = \frac{{T_2}}{{T_1}}$$

Solving for $T_1$:

$$T_1 = \frac{{T_2}}{{1.004}}$$

$$T_1 \approx \frac{{T_2}}{{1.004}}$$

$$T_1 \approx \frac{{1 \, K}}{{1.004}}$$

$$T_1 \approx 0.996 \, K$$

So, the initial temperature of the gas was approximately $0.996 \, K$.

For a certain gas in a closed container, the pressure has been raised by #0.4%#, and the temperature was raised by #"1 K"#. What temperature did the gas start at?

#a)# #"250 K"# #b)# #"200 K"# #c)# #"298 K"# #d)# #"300 K"#

#a)# #"250 K"#
#b)# #"200 K"#
#c)# #"298 K"#
#d)# #"300 K"#