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3 years ago

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Tricia puts 44 g of dry ice (solid CO2) into a 2.0-L container and seals the top. The dry ice turns to gas at room temperature (

20°C) Find the pressure increase in the 2.0-L container. (One mole of CO2 has a mass of 44 g, R = 0.0821 L×atm/molxK. Ignore the initial volume of the dry ice.)
A) 6 atm
B) 18 atm
C) 2 atm
D) 12 atm

1 answer:

Irina-Kira [14]3 years ago

6 0

Answer:

Corect answer is D

Explanation:

Assuming that the C O 2 gas is behaving ideally, therefore, we can use the ideal gas law to find the pressure increase in the container by:

P V=nRT ⇒ P=n R T /V

n=no of moles of the gas = mass/molar mass

Molar mass o f C O 2=44g/mol, mass = 44g

mole n = 1mole

T=20C=293K

R=0.0821L.atm/mol.K

P=nRT/V

P = 1 x 0.0821 x 293/2

P = 12atm

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2 years ago

A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m.

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$D=99.4665307m \approx 99.5m$

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From the question we are told that

Mass $m=6kg$

Velocity of mass $V_m=16$

Force of Tunnel $F_t=8N$

Length of Tunnel $L_t=1.6$

Height of frictional incline $H_i=2.9$

Angle of inclination $\angle =16 \textdegree$

Acceleration due to gravity $g=9.8m/s^2$

First Frictional surface has a coeﬃcient $\alpha_1 =0.21\ for\ d_c=1$

Second Frictional surface has a coeﬃcient $\alpha _2=0.1$

Generally the initial Kinetic energy is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}(6)(16)^2$

$K.E=768$

Generally the work done by the Tunnel is mathematically given as

$w_t=F_t*d_t$

$w_t=8*1.6$

$w_t=12.8J$

Therefore

$Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t$

$E_t=K.E+E_t\\E_t=768J+12.8J$

$E_t=780.8J$

Generally the energy lost while climbing is mathematically given as

$E_c=mgh$

$E_c=(6)(9.8)(2.9)$

$E_c=170.52J$

Generally the energy lost to friction is mathematically given as

$E_f=\alpha *m*g*cos\textdegree*d_c$

$E_f=0.21*6*9.8*cos16*1$

$E_f=11.86965942 \approx 12J$

Generally the energy left in the form of mass $Em$ is mathematically given as

$E_m=E_t+E_c+E_f$

$E_m=(768J)-(170.52)-(12)$

$E_m=585.48J$

Since

$E_m=\alpha_2*g*m*d$

Therefore

It slide along the second frictional region

$D=\frac{585.46}{0.1*9.81*6}$

$D=99.4665307m \approx 99.5m$

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3 years ago