371547
An ideal gas has volume \(V_{0}\) at \(27^\circ C\). It is heated at constant pressure so that its volume becomes \(2 V_{0}\). The final temperature is
371548
When heat is given to a gas in an isobaric process, then
1 Internal energy of the gas increases
2 The work is done by the gas
3 Internal energy of the gas decreases
4 Both (1) and (2)
Explanation:
When heat is supplied at constant pressure, a part of it goes in the expansion of gas and remaining part is used to increase the temperature of the gas which in turn increases the internal energy.
PHXI12:THERMODYNAMICS
371549
The given figure represents two isobaric processes for the same mass of an ideal gas, then
1 \({P_{1}=P_{2}}\)
2 \({P_{2} \geq P_{1}}\)
3 \({P_{2}>P_{1}}\)
4 \({P_{1}>P_{2}}\)
Explanation:
As the given processes are isobaric, the reciprocal of the slope of processes will represent the value of pressure. \({P V=n R T}\) \({V=\left(\dfrac{n R}{P}\right) T}\) \({\therefore}\) Slope \({=\dfrac{n R}{P}}\) As, slope of \({P_{2}>}\) slope of \({P_{1}}\) \({\therefore \quad P_{1}>P_{2}}\) So, correct option is (4).
JEE - 2024
PHXI12:THERMODYNAMICS
371550
One mole of \({O_2}\) gas is heated at constant pressure starting at \(27^\circ C\). How much energy must be added to the gas as heat to double its volume?
1 \(750\,R\)
2 \({\rm{Zero}}\)
3 \(1050\,R\)
4 \(450\,R\)
Explanation:
Here, \({V_1} = V,\,\,{V_2} = 2\;V\) \({T_1} = 27^\circ C = 300\;K\) At constant pressure \(\frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}}\quad \therefore {T_2} = \left( {\frac{{{V_2}}}{{{V_1}}}} \right) \times {T_1} = 600\;K\) Heat absorbed by the gas at constant pressure \( = n{C_p}\Delta T = 1 \times \frac{7}{2}R \times 300 = 1050R.\)
371547
An ideal gas has volume \(V_{0}\) at \(27^\circ C\). It is heated at constant pressure so that its volume becomes \(2 V_{0}\). The final temperature is
371548
When heat is given to a gas in an isobaric process, then
1 Internal energy of the gas increases
2 The work is done by the gas
3 Internal energy of the gas decreases
4 Both (1) and (2)
Explanation:
When heat is supplied at constant pressure, a part of it goes in the expansion of gas and remaining part is used to increase the temperature of the gas which in turn increases the internal energy.
PHXI12:THERMODYNAMICS
371549
The given figure represents two isobaric processes for the same mass of an ideal gas, then
1 \({P_{1}=P_{2}}\)
2 \({P_{2} \geq P_{1}}\)
3 \({P_{2}>P_{1}}\)
4 \({P_{1}>P_{2}}\)
Explanation:
As the given processes are isobaric, the reciprocal of the slope of processes will represent the value of pressure. \({P V=n R T}\) \({V=\left(\dfrac{n R}{P}\right) T}\) \({\therefore}\) Slope \({=\dfrac{n R}{P}}\) As, slope of \({P_{2}>}\) slope of \({P_{1}}\) \({\therefore \quad P_{1}>P_{2}}\) So, correct option is (4).
JEE - 2024
PHXI12:THERMODYNAMICS
371550
One mole of \({O_2}\) gas is heated at constant pressure starting at \(27^\circ C\). How much energy must be added to the gas as heat to double its volume?
1 \(750\,R\)
2 \({\rm{Zero}}\)
3 \(1050\,R\)
4 \(450\,R\)
Explanation:
Here, \({V_1} = V,\,\,{V_2} = 2\;V\) \({T_1} = 27^\circ C = 300\;K\) At constant pressure \(\frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}}\quad \therefore {T_2} = \left( {\frac{{{V_2}}}{{{V_1}}}} \right) \times {T_1} = 600\;K\) Heat absorbed by the gas at constant pressure \( = n{C_p}\Delta T = 1 \times \frac{7}{2}R \times 300 = 1050R.\)
371547
An ideal gas has volume \(V_{0}\) at \(27^\circ C\). It is heated at constant pressure so that its volume becomes \(2 V_{0}\). The final temperature is
371548
When heat is given to a gas in an isobaric process, then
1 Internal energy of the gas increases
2 The work is done by the gas
3 Internal energy of the gas decreases
4 Both (1) and (2)
Explanation:
When heat is supplied at constant pressure, a part of it goes in the expansion of gas and remaining part is used to increase the temperature of the gas which in turn increases the internal energy.
PHXI12:THERMODYNAMICS
371549
The given figure represents two isobaric processes for the same mass of an ideal gas, then
1 \({P_{1}=P_{2}}\)
2 \({P_{2} \geq P_{1}}\)
3 \({P_{2}>P_{1}}\)
4 \({P_{1}>P_{2}}\)
Explanation:
As the given processes are isobaric, the reciprocal of the slope of processes will represent the value of pressure. \({P V=n R T}\) \({V=\left(\dfrac{n R}{P}\right) T}\) \({\therefore}\) Slope \({=\dfrac{n R}{P}}\) As, slope of \({P_{2}>}\) slope of \({P_{1}}\) \({\therefore \quad P_{1}>P_{2}}\) So, correct option is (4).
JEE - 2024
PHXI12:THERMODYNAMICS
371550
One mole of \({O_2}\) gas is heated at constant pressure starting at \(27^\circ C\). How much energy must be added to the gas as heat to double its volume?
1 \(750\,R\)
2 \({\rm{Zero}}\)
3 \(1050\,R\)
4 \(450\,R\)
Explanation:
Here, \({V_1} = V,\,\,{V_2} = 2\;V\) \({T_1} = 27^\circ C = 300\;K\) At constant pressure \(\frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}}\quad \therefore {T_2} = \left( {\frac{{{V_2}}}{{{V_1}}}} \right) \times {T_1} = 600\;K\) Heat absorbed by the gas at constant pressure \( = n{C_p}\Delta T = 1 \times \frac{7}{2}R \times 300 = 1050R.\)
371547
An ideal gas has volume \(V_{0}\) at \(27^\circ C\). It is heated at constant pressure so that its volume becomes \(2 V_{0}\). The final temperature is
371548
When heat is given to a gas in an isobaric process, then
1 Internal energy of the gas increases
2 The work is done by the gas
3 Internal energy of the gas decreases
4 Both (1) and (2)
Explanation:
When heat is supplied at constant pressure, a part of it goes in the expansion of gas and remaining part is used to increase the temperature of the gas which in turn increases the internal energy.
PHXI12:THERMODYNAMICS
371549
The given figure represents two isobaric processes for the same mass of an ideal gas, then
1 \({P_{1}=P_{2}}\)
2 \({P_{2} \geq P_{1}}\)
3 \({P_{2}>P_{1}}\)
4 \({P_{1}>P_{2}}\)
Explanation:
As the given processes are isobaric, the reciprocal of the slope of processes will represent the value of pressure. \({P V=n R T}\) \({V=\left(\dfrac{n R}{P}\right) T}\) \({\therefore}\) Slope \({=\dfrac{n R}{P}}\) As, slope of \({P_{2}>}\) slope of \({P_{1}}\) \({\therefore \quad P_{1}>P_{2}}\) So, correct option is (4).
JEE - 2024
PHXI12:THERMODYNAMICS
371550
One mole of \({O_2}\) gas is heated at constant pressure starting at \(27^\circ C\). How much energy must be added to the gas as heat to double its volume?
1 \(750\,R\)
2 \({\rm{Zero}}\)
3 \(1050\,R\)
4 \(450\,R\)
Explanation:
Here, \({V_1} = V,\,\,{V_2} = 2\;V\) \({T_1} = 27^\circ C = 300\;K\) At constant pressure \(\frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}}\quad \therefore {T_2} = \left( {\frac{{{V_2}}}{{{V_1}}}} \right) \times {T_1} = 600\;K\) Heat absorbed by the gas at constant pressure \( = n{C_p}\Delta T = 1 \times \frac{7}{2}R \times 300 = 1050R.\)
