Answer:
184.62 ml
Explanation:
Let
and
be the initial and
and
be the final pressure, volume, and temperature of the gas respectively.
Given that the pressure remains constant, so
...(i)
= 200 ml
K
K
From the ideal gas equation, pv=mRT
Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.
For the initial condition,
![p_1v_1=mRT_1 \\\\mR= \frac{p_1v_1}{T_1}\cdots(ii)](https://tex.z-dn.net/?f=p_1v_1%3DmRT_1%20%5C%5C%5C%5CmR%3D%20%5Cfrac%7Bp_1v_1%7D%7BT_1%7D%5Ccdots%28ii%29)
For the final condition,
![p_2v_2=mRT_2 \\\\mR= \frac{p_2v_2}{T_2}\cdots(iii)](https://tex.z-dn.net/?f=p_2v_2%3DmRT_2%20%5C%5C%5C%5CmR%3D%20%5Cfrac%7Bp_2v_2%7D%7BT_2%7D%5Ccdots%28iii%29)
Equating equation (i), and (ii)
![\frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7Bp_1v_1%7D%7BT_1%7D%3D%5Cfrac%7Bp_2v_2%7D%7BT_2%7D)
[from equation (i)]
![v_2=\frac{T_2}{T_1} \times v_1](https://tex.z-dn.net/?f=v_2%3D%5Cfrac%7BT_2%7D%7BT_1%7D%20%5Ctimes%20v_1)
Putting all the given values, we have
![v_2=\frac{276}{299} \times 200 = 184.62 \; ml](https://tex.z-dn.net/?f=v_2%3D%5Cfrac%7B276%7D%7B299%7D%20%5Ctimes%20200%20%3D%20184.62%20%5C%3B%20ml)
Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.