Answer:
The new pressure will be 0.225 kPa.
Explanation:
Applying combined gas law:
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume at initial temperature [tex]T_1[/tex].
[tex]P_2\text{ and }V_2[/tex] are final pressure and volume at initial temperature [tex]T_2[/tex].
We are given:
[tex]P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V[/tex]
[tex]T_1=-25^oC=248.15 K[/tex]
[tex]T _2 = 16^oC=289.15 K[/tex]
Putting values in above equation, we get:
[tex]\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}[/tex]
[tex]P_1=0.225 kPa[/tex]
Hence, the new pressure will be 0.225 kPa.
