A given mass of gas occupies a volume of 200 ml at 37°C. To what temperature must the gas be heated to make its final volume as 500 ml. Assume that the pressure of the gas remains constant.
Answers
Let,
By Charles Law
Substituting the value,
Therefore, the final temperature of the gas is 502°C
Know more about Charles Law.
Charles's law is an experimental gas law that describes how gases tend to expand when heated. A modern statement of Charles's law is: When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion.
Answer:
\huge{ \underline{ \bf{Given \: : }}}
Given:
Let,
\tt{v_1 = 200ml}v
1
=200ml
\tt{t_2 = 37 {}^{0} c} \: or \: (273 + 37)k = 310kt
2
=37
0
cor(273+37)k=310k
\tt \: v_2 = 500mlv
2
=500ml
\tt \: t_2 = ?t
2
=?
\huge{ \underline{ \underline{ \bf{Solution \: : - }}}}
Solution:−
By Charles Law
\color{purple} \leadsto \red {\boxed{ \tt{ \green{ \frac{v _1 }{t _1} = \frac{v _ 2}{t _ 2} }}}}⇝
t
1
v
1
=
t
2
v
2
Substituting the value,
\tt \to \frac{200}{310} = \frac{500}{t _2 }→
310
200
=
t
2
500
\tt \to \: t_2 = \frac{5 \cancel{00} \times \cancel{310}}{ \cancel2 \cancel{00} }→t
2
=
2
00
5
00
×
310
\tt \to \: t_2 = 775k→t
2
=775k
\because \boxed{ \tt{ \blue{1 {}^{0} c = 272k}}}∵
1
0
c=272k
\tt \to \: t_2 = (775 - 273) {}^{0} = 502 {}^{0} c→t
2
=(775−273)
0
=502
0
c
\tt \mid \: \color{gold} {t_2 = 502 {}^{0} c} \mid∣t
2
=502
0
c∣
Therefore, the final temperature of the gas is 502°C
Know more about Charles Law.
Charles's law is an experimental gas law that describes how gases tend to expand when heated. A modern statement of Charles's law is: When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion.
Answer:
\huge{ \underline{ \bf{Given \: : }}}
Given:
Let,
\tt{v_1 = 200ml}v
1
=200ml
\tt{t_2 = 37 {}^{0} c} \: or \: (273 + 37)k = 310kt
2
=37
0
cor(273+37)k=310k
\tt \: v_2 = 500mlv
2
=500ml
\tt \: t_2 = ?t
2
=?
\huge{ \underline{ \underline{ \bf{Solution \: : - }}}}
Solution:−
By Charles Law
\color{purple} \leadsto \red {\boxed{ \tt{ \green{ \frac{v _1 }{t _1} = \frac{v _ 2}{t _ 2} }}}}⇝
t
1
v
1
=
t
2
v
2
Substituting the value,
\tt \to \frac{200}{310} = \frac{500}{t _2 }→
310
200
=
t
2
500
\tt \to \: t_2 = \frac{5 \cancel{00} \times \cancel{310}}{ \cancel2 \cancel{00} }→t
2
=
2
00
5
00
×
310
\tt \to \: t_2 = 775k→t
2
=775k
\because \boxed{ \tt{ \blue{1 {}^{0} c = 272k}}}∵
1
0
c=272k
\tt \to \: t_2 = (775 - 273) {}^{0} = 502 {}^{0} c→t
2
=(775−273)
0
=502
0
c
\tt \mid \: \color{gold} {t_2 = 502 {}^{0} c} \mid∣t
2
=502
0
c∣
Therefore, the final temperature of the gas is 502°C
Know more about Charles Law.
Charles's law is an experimental gas law that describes how gases tend to expand when heated. A modern statement of Charles's law is: When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion.
Answer:
\huge{ \underline{ \bf{Given \: : }}}
Given:
Let,
\tt{v_1 = 200ml}v
1
=200ml
\tt{t_2 = 37 {}^{0} c} \: or \: (273 + 37)k = 310kt
2
=37
0
cor(273+37)k=310k
\tt \: v_2 = 500mlv
2
=500ml
\tt \: t_2 = ?t
2
=?
\huge{ \underline{ \underline{ \bf{Solution \: : - }}}}
Solution:−
By Charles Law
\color{purple} \leadsto \red {\boxed{ \tt{ \green{ \frac{v _1 }{t _1} = \frac{v _ 2}{t _ 2} }}}}⇝
t
1
v
1
=
t
2
v
2
Substituting the value,
\tt \to \frac{200}{310} = \frac{500}{t _2 }→
310
200
=
t
2
500
\tt \to \: t_2 = \frac{5 \cancel{00} \times \cancel{310}}{ \cancel2 \cancel{00} }→t
2
=
2
00
5
00
×
310
\tt \to \: t_2 = 775k→t
2
=775k
\because \boxed{ \tt{ \blue{1 {}^{0} c = 272k}}}∵
1
0
c=272k
\tt \to \: t_2 = (775 - 273) {}^{0} = 502 {}^{0} c→t
2
=(775−273)
0
=502
0
c
\tt \mid \: \color{gold} {t_2 = 502 {}^{0} c} \mid∣t
2
=502
0
c∣
Therefore, the final temperature of the gas is 502°C
Know more about Charles Law.
Charles's law is an experimental gas law that describes how gases tend to expand when heated. A modern statement of Charles's law is: When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion.
Answer:
\huge{ \underline{ \bf{Given \: : }}}
Given:
Let,
\tt{v_1 = 200ml}v
1
=200ml
\tt{t_2 = 37 {}^{0} c} \: or \: (273 + 37)k = 310kt
2
=37
0
cor(273+37)k=310k
\tt \: v_2 = 500mlv
2
=500ml
\tt \: t_2 = ?t
2
=?
\huge{ \underline{ \underline{ \bf{Solution \: : - }}}}
Solution:−
By Charles Law
\color{purple} \leadsto \red {\boxed{ \tt{ \green{ \frac{v _1 }{t _1} = \frac{v _ 2}{t _ 2} }}}}⇝
t
1
v
1
=
t
2
v
2
Substituting the value,
\tt \to \frac{200}{310} = \frac{500}{t _2 }→
310
200
=
t
2
500
\tt \to \: t_2 = \frac{5 \cancel{00} \times \cancel{310}}{ \cancel2 \cancel{00} }→t
2
=
2
00
5
00
×
310
\tt \to \: t_2 = 775k→t
2
=775k
\because \boxed{ \tt{ \blue{1 {}^{0} c = 272k}}}∵
1
0
c=272k
\tt \to \: t_2 = (775 - 273) {}^{0} = 502 {}^{0} c→t
2
=(775−273)
0
=502
0
c
\tt \mid \: \color{gold} {t_2 = 502 {}^{0} c} \mid∣t
2
=502
0
c∣
Therefore, the final temperature of the gas is 502°C
Know more about Charles Law.
Charles's law is an experimental gas law that describes how gases tend to expand when heated. A modern statement of Charles's law is: When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion.