Ask question

Ask question

All categories

2 years ago

12

At what temperature, in Kelvin, will a 250 milliliter cylinder containing 0.100 moles of helium gas have a pressure of 2.50 atmo

sphere? Write your answer as a decimal number to three significant figures. Given: R = 0.08205 liter · atmosphere/mole · Kelvin

2 answers:

lozanna [386]2 years ago

7 0

Use the ideal gas formula-----> PV= nRT

P= 2.50 atm
V= 250 mL= 0.250 L
n= 0.100 moles
R= 0.0821 atmxL/molesxK
T= ?

T= PV/nR

T= (2.50 atm x 0.250 L) / (0.100 moles x 0.0821)= 76.1 K

Gnom [1K]2 years ago

4 0

The correct answer is 72.6

You might be interested in

How many moles of gas does it take to occupy 120 L at a pressure of 233 kPa and a temperature of 340 K?

natka813 [3]

A would be correct have a nice day.

5 0

2 years ago

Read 2 more answers

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

1 year ago

Why is it usually warmer at Earth’s equator than at its poles?

MissTica

Answer:

the equator is closer to the sun

8 0

2 years ago

Which is smaller 25 mm or 25cm

umka21 [38]

Answer:

25 mm is smaller.

5 0

2 years ago

What is 25 m rounded to one significant figure

den301095 [7]

Rounded to 1 significant figure, 25 m would go to 30. This is because 0 isn't significant, so the 3 is the only significant figure.

8 0

2 years ago

Read 2 more answers