Answer:
[tex]\boxed{\text{25. 20 L; 26. 49 K}}[/tex]
Explanation:
25. Boyle's Law
The temperature and amount of gas are constant, so we can use Boyle’s Law.
[tex]p_{1}V_{1} = p_{2}V_{2}[/tex]
Data:
[tex]\begin{array}{rcrrcl}p_{1}& =& \text{100 kPa}\qquad & V_{1} &= & \text{10.00 L} \\p_{2}& =& \text{50 kPa}\qquad & V_{2} &= & ?\\\end{array}[/tex]
Calculations:
[tex]\begin{array}{rcl}100 \times 10.00 & =& 50V_{2}\\1000 & = & 50V_{2}\\V_{2} & = &\textbf{20 L}\\\end{array}\\\text{The new volume will be } \boxed{\textbf{20 L}}[/tex]
26. Ideal Gas Law
We have p, V and n, so we can use the Ideal Gas Law to calculate the volume.
pV = nRT
Data:
p = 101.3 kPa
V = 20 L
n = 5 mol
R = 8.314 kPa·L·K⁻¹mol⁻¹
Calculation:
101.3 × 20 = 5 × 8.314 × T
2026 = 41.57T
[tex]T = \dfrac{2026}{41.57} = \textbf{49 K}\\\\\text{The Kelvin temperature is }\boxed{\textbf{49 K}}[/tex]
