Work Done on Compressing a Gas NEET Questions

Work Done on Compressing a Gas MCQ Questions

7.
Which of the following correctly explains why a gas heats up when compressed rapidly in a thermally insulated cylinder?
A.
Compression reduces molecular mass
B.
The gas molecules absorb infrared radiation from the walls
C.
Heat flows in from the hot piston
D.
The piston moves towards the gas molecules; each molecule colliding with the moving piston rebounds faster โ€” gaining kinetic energy, raising temperature
ANSWER :
D. The piston moves towards the gas molecules; each molecule colliding with the moving piston rebounds faster โ€” gaining kinetic energy, raising temperature
8.
In NCERT Exercise 11.5, a gas is taken adiabatically from state A to B with work done ON the system = 22.3 J. The change in internal energy ฮ”U is:
A.
ฮ”U = 0 (adiabatic means no energy change)
B.
ฮ”U = +22.3 J (work done on gas increases internal energy)
C.
ฮ”U = โˆ’22.3 J (internal energy decreases)
D.
ฮ”U = +44.6 J
ANSWER :
B. ฮ”U = +22.3 J (work done on gas increases internal energy)
9.
Work done BY gas during isothermal compression from Vโ‚ to Vโ‚‚ (Vโ‚‚ < Vโ‚) at temperature T for n moles is:
A.
W = nRT ln(Vโ‚/Vโ‚‚) โ€” positive for compression
B.
W = nR(Tโ‚ โˆ’ Tโ‚‚)/(ฮณโˆ’1)
C.
W = 0 (no temperature change)
D.
W = nRT ln(Vโ‚‚/Vโ‚) โ€” negative for compression
ANSWER :
D. W = nRT ln(Vโ‚‚/Vโ‚) โ€” negative for compression
10.
During isothermal compression of an ideal gas, ฮ”U = 0. The First Law gives:
A.
ฮ”Q = 0 (no heat exchange in isothermal)
B.
ฮ”Q = โˆ’ฮ”U = 0
C.
ฮ”Q = ฮ”W > 0 (heat absorbed during compression)
D.
ฮ”Q = ฮ”W < 0 (heat released by gas to reservoir during compression)
ANSWER :
D. ฮ”Q = ฮ”W < 0 (heat released by gas to reservoir during compression)
11.
For isothermal compression of 1 mole of ideal gas at T = 300 K from Vโ‚ to Vโ‚/2 (volume halved). Work done BY gas is: (R = 8.3 J/molยทK, ln 2 = 0.693)
A.
0 J
B.
+863 J
C.
โˆ’1726 J
D.
+1726 J
ANSWER :
C. โˆ’1726 J
12.
For isothermal compression of n moles of ideal gas at temperature T, if the volume is reduced to one-quarter (Vโ‚‚ = Vโ‚/4), the work done BY the gas is:
A.
W = โˆ’4nRT
B.
W = nRT/4
C.
W = nRT ln(1/4) = โˆ’2nRT ln 2
D.
W = nRT ln 4 = +2nRT ln 2
ANSWER :
C. W = nRT ln(1/4) = โˆ’2nRT ln 2