Question

A bicycle tire has a pressure of 6.9 × 105 Pa at a temperature of 17.5°C and contains 2.00 L of gas. show answer Incorrect Answer What will its pressure be, in pascals, if you let out an amount of air that has a volume of 95 cm3 at atmospheric pressure and at the temperature of the tire? Assume tire temperature and volume remain constant. P2 = 14526315.79| sin() cos() tan() cotan() asin() acos() atan() acotan() sinh() cosh() tanh() cotanh() Degrees Radians π ( ) 7 8 9 HOME E ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + - 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 8% Potential 92% Submissions Attempts remaining: 1 (4% per attempt) detailed view 1 4% 2 4%

Answer 1

Answer:

the final pressure in pascal = $6.30*10^5 \ \ Pa$

Explanation:

Given that:

initial pressure = $6.9*10^5 \ Pa$

Initial Temperature = 17.5 °C  = (17.5 + 273) K = 290.5 K

Volume = 2.00 L = 2.00 × 10⁻³ m³

Initial volume of the air occupied by gas V' = 95 cm³ = 95× 10⁻⁶ m

Using the ideal gas temperature;

PV = nKT

where

K = Boltzmann constant = 1.38 × 10⁻²³

From above expression;

$n = \frac{PV}{KT}$

$n = \frac{6.9*10^5*2.00*10^{-3}}{1.38*10^{-23}*290.5}$

$n = 3.44*10^{23}$

The number of moles that were removed from the tire is calculated as

$\Delta \ n = \frac{P_{atm}*V'}{KT}$

where

$P_{atm}$ = atmospheric pressure = $1.013*10^5$

$\Delta \ n = \frac{1.013*10^5*95*10^{-6}}{1.38*10^{-23}*290.5}$

$\Delta \ n = 2.4*10^{21}$

The remaining number of moles after the release of gas is

$n_2 = n- \Delta n$

$n_2 = 3.44*10^{23} -2.4*10^{21}$

$n_2 = 3.416*10^{23}$

Using the expression  $P_2 = \frac{n_2 KT}{V}$ to determine the final pressure:

$P_2 = \frac{3.416*10^{23}*1.38*10^{-23}*290.5 }{2.00*10^{-3}}$

$P_2 = 6.30*10^5 \ \ Pa$

Hence, the final pressure in pascal = $6.30*10^5 \ \ Pa$