Question

for many purposes we can treat dimethyl ether (c₂h₆o) as an ideal gas at temperatures above its boiling point of -24. °c.
suppose the pressure on a 850. ml sample of dimethyl ether gas at -14.0 °c is reduced to one - third its initial value.
is it possible to change the temperature of the dimethyl ether at the same time such that the volume of the gas doesnt change?
if you answered yes, calculate the new temperature of the gas. round your answer to the nearest °c.

Explanation:

Step1: Convert initial temperature to Kelvin

The initial temperature \(T_1=-14.0^{\circ}C\). Using the conversion \(T(K)=T(^{\circ}C)+273.15\), we have \(T_1 = (-14.0)+273.15=259.15\space K\).

Step2: Apply the ideal gas law (since \(V\) is constant, \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\))

Let \(P_1 = P\), then \(P_2=\frac{1}{3}P\). Substituting into \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\), we get \(\frac{P}{259.15}=\frac{\frac{1}{3}P}{T_2}\). The \(P\) cancels out, and we can solve for \(T_2\). Cross - multiplying gives \(T_2=\frac{259.15}{3}\approx86.38\space K\).

Step3: Convert the new temperature to Celsius

Using \(T(^{\circ}C)=T(K)-273.15\), we have \(T_2=(86.38 - 273.15)^{\circ}C=-187^{\circ}C\)

Answer:

yes, \(-187^{\circ}C\)